If a point $P(\alpha, \beta, \gamma)$ satisfying $(\alpha \beta \gamma) \begin{pmatrix} 2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8 \end{pmatrix} = (0, 0, 0)$ lies on the plane $2x + 4y + 3z = 5$, then $6\alpha + 9\beta + 7\gamma$ is equal to: Options: (A) -1 (B) 11/5 (C) 5/4 (D) 11

### Analyzing the Matrix Setup We begin with the matrix equation: $(\alpha, \beta, \gamma) \begin{pmatrix} 2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8 \end{pmatrix} = (0, 0, 0)$ This matrix expression represents a system of three linear equations. By multiplying the row vector by each column of the matrix, we obtain: $$2\alpha + 9\beta + 8\gamma = 0$$ $$10\alpha + 3\beta + 4\gamma = 0$$ $$8\alpha + 8\beta + 8\gamma = 0$$ ### The Power of Simplification Observe the third equation: $8\alpha + 8\beta + 8\gamma = 0$. Dividing by $8$ yields the simplified relation: $$\alpha + \beta + \gamma = 0 \quad \Rightarrow \quad \gamma = -\alpha - \beta$$ Substitute this expression for $\gamma$ into the first equation: $$2\alpha + 9\beta + 8(-\alpha - \beta) = 0$$ Expanding and simplifying the terms: $$2\alpha + 9\beta - 8\alpha - 8\beta = 0$$ $$-6\alpha + \beta = 0 \quad \Rightarrow \quad \beta = 6\alpha$$ Using the relation $\gamma = -\alpha - \beta$, we find $\gamma$ in terms of $\alpha$: $$\gamma = -\alpha - 6\alpha = -7\alpha$$ Thus, the point $P$ is parameterized as $(\alpha, 6\alpha, -7\alpha)$. ### The Geometric Constraint The point $P$ lies on the plane defined by $2x + 4y + 3z = 5$. Substituting our parameterized coordinates into this plane equation gives: $$2(\alpha) + 4(6\alpha) + 3(-7\alpha) = 5$$ Performing the arithmetic: $$2\alpha + 24\alpha - 21\alpha = 5$$ $$5\alpha = 5 \quad \Rightarrow \quad \alpha = 1$$ With $\alpha = 1$, we determine the specific coordinates: $$\beta = 6(1) = 6$$ $$\gamma = -7(1) = -7$$ The point $P$ is $(1, 6, -7)$. ### The Final Calculation The problem requires the evaluation of the linear combination $6\alpha + 9\beta + 7\gamma$. Substituting our found values: $$6(1) + 9(6) + 7(-7)$$ $$6 + 54 - 49$$ $$60 - 49 = 11$$ The final result is **11**.

Why do we express everything in terms of $\alpha$?

In a homogeneous system of linear equations where the determinant of the matrix is zero, there isn't a unique solution like $(0, 0, 0)$ for all variables unless specified. Instead, there is a line of solutions. By expressing $\beta$ and $\gamma$ in terms of $\alpha$, we are essentially finding the direction vector of this line. This allows us to treat the point $P$ as a single variable problem, which makes the final step with the plane equation much easier!

JEE Main 2023 (31 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: If a point $P (α, β, γ)$ satisfying $(α β γ) 2981034888 = (0, 0, 0)$ lies on the plane $2 x + 4 y + 3 z = 5$ , then $6 α + 9 β + 7 γ$ is equal to:

Select Answer:

Visualized Solution

Setting up the Matrix Equation

Given matrix equation:
$(α, β, γ) 2981034888 = (0, 0, 0)$

First Linear Equation

Multiplying the row vector by the first column:
$2 α + 9 β + 8 γ = 0$

Second Linear Equation

Multiplying the row vector by the second column:
$10 α + 3 β + 4 γ = 0$

Third Linear Equation

Multiplying the row vector by the third column:
$8 α + 8 β + 8 γ = 0$
Dividing by $8$ :
$α + β + γ = 0$

Solving the System

From the third equation:
$γ = - α - β$
Substitute into the first equation:
$2 α + 9 β + 8 (- α - β) = 0$
$⟹ β = 6 α$
Substitute $β$ back to find $γ$ :
$γ = - α - 6 α = - 7 α$

The Plane Equation

Point $P (α, β, γ)$ lies on the plane:
$2 x + 4 y + 3 z = 5$

Substituting into Plane Equation

Substitute the coordinates of $P$ into the plane equation:
$2 α + 4 β + 3 γ = 5$

Calculating $α$

Substitute $β = 6 α$ and $γ = - 7 α$ :
$2 α + 4 (6 α) + 3 (- 7 α) = 5$
$2 α + 24 α - 21 α = 5$
$5 α = 5 ⟹ α = 1$

Coordinates of Point $P$

Now find $β$ and $γ$ :
$β = 6 (1) = 6$
$γ = - 7 (1) = - 7$
Coordinates of $P$ are $(1, 6, - 7)$

Evaluating the Expression

We need to evaluate:
$6 α + 9 β + 7 γ$
Substitute the values:
$6 (1) + 9 (6) + 7 (- 7)$

Final Result

Calculate the final sum:
$6 + 54 - 49$
$= 60 - 49 = 11$
Final Answer: $11$

00:00 / 00:00

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Solution Diagram

Analyzing the Matrix Setup

We begin with the matrix equation:

(α, β, γ) 2981034888 = (0, 0, 0)

This matrix expression represents a system of three linear equations. By multiplying the row vector by each column of the matrix, we obtain:

2 α + 9 β + 8 γ = 0

10 α + 3 β + 4 γ = 0

8 α + 8 β + 8 γ = 0

The Power of Simplification

Observe the third equation:

8 α + 8 β + 8 γ = 0

. Dividing by

8

yields the simplified relation:

α + β + γ = 0 \Rightarrow γ = - α - β

Substitute this expression for

γ

into the first equation:

2 α + 9 β + 8 (- α - β) = 0

Expanding and simplifying the terms:

2 α + 9 β - 8 α - 8 β = 0

- 6 α + β = 0 \Rightarrow β = 6 α

Using the relation

γ = - α - β

, we find

γ

in terms of

α

γ = - α - 6 α = - 7 α

Thus, the point

P

is parameterized as

(α, 6 α, - 7 α)

The Geometric Constraint

The point

P

lies on the plane defined by

2 x + 4 y + 3 z = 5

. Substituting our parameterized coordinates into this plane equation gives:

2 (α) + 4 (6 α) + 3 (- 7 α) = 5

Performing the arithmetic:

2 α + 24 α - 21 α = 5

5 α = 5 \Rightarrow α = 1

With

α = 1

, we determine the specific coordinates:

β = 6 (1) = 6

γ = - 7 (1) = - 7

The point

P

(1, 6, - 7)

The Final Calculation

The problem requires the evaluation of the linear combination

6 α + 9 β + 7 γ

. Substituting our found values:

6 (1) + 9 (6) + 7 (- 7)

6 + 54 - 49

60 - 49 = 11

The final result is 11.

Common Traps & FAQs

Wait, why are there three equations? Can't we just solve for one?

That is a great question! When you multiply a row vector by a

3 \times 3

matrix, the result is a row vector of length 3. Since the result is

(0, 0, 0)

, each component of that resulting vector must be zero. This naturally gives you three distinct linear equations. You need all three to fully constrain the relationship between

α, β, and γ

Why do we express everything in terms of

α

In a homogeneous system of linear equations where the determinant of the matrix is zero, there isn't a unique solution like

(0, 0, 0)

for all variables unless specified. Instead, there is a line of solutions. By expressing

β

and

γ

in terms of

α

, we are essentially finding the direction vector of this line. This allows us to treat the point

P

as a single variable problem, which makes the final step with the plane equation much easier!

What if the system had no solution?

In this specific problem, the system is consistent because it is homogeneous (all equations equal zero). A homogeneous system always has at least the trivial solution

(0, 0, 0)

. If the plane equation had been something like

0 = 5

, then there would be no solution. But here, the geometry guarantees that the line defined by the matrix equations intersects the plane, giving us a unique point

P

Animated Solution for Mathematics - Matrices and Determinants: If a point $P (α, β, γ)$ satisfying $(α β γ) 2981034888 = (0, 0, 0)$ lies on the plane $2 x + 4 y + 3 z = 5$ , then $6 α + 9 β + 7 γ$ is equal to:

Select Answer:

Visualized Solution

Setting up the Matrix Equation

First Linear Equation

Second Linear Equation

Third Linear Equation

Solving the System

The Plane Equation

Substituting into Plane Equation

Calculating $α$

Coordinates of Point $P$

Evaluating the Expression

Final Result

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Matrix Setup

The Power of Simplification

The Geometric Constraint

The Final Calculation

Similar Questions

If for some $α$ and $β$ in $R$ , the intersection of the following three planes $x + 4 y - 2 z = 1$ , $x + 7 y - 5 z = β$ , $x + 5 y + α z = 5$ is a line in $R^{3}$ , then $α + β$ is equal to:

Comprehension Passage

If the point $P (a, b, c)$ , with reference to (E), lies on the plane $2 x + y + z = 1$ , then the value of $7 a + b + c$ is

Let $ω$ be a solution of $x^{3} - 1 = 0$ with $Im (ω) > 0$ , if $a = 2$ with $b$ and $c$ satisfying (E), then the value of $\frac{3}{ω ^{a}} + \frac{1}{ω ^{b}} + \frac{3}{ω ^{c}}$ is equal to

Let $b = 6$ , with $a$ and $c$ satisfying (E). If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ , then $\sum_{n = 0}^{\infty} (\frac{1}{α} + \frac{1}{β})^{n}$ is

Let the system of linear equations $- x + 2 y - 9 z = 7$ , $- x + 3 y + 7 z = 9$ , $- 2 x + y + 5 z = 8$ , $- 3 x + y + 13 z = λ$ has a unique solution $x = α, y = β, z = γ$ . Then the distance of the point $(α, β, γ)$ from the plane $2 x -2 y + z = λ$ is

If the system of linear equations $8 x + y + 4 z = - 2$ , $x + y + z = 0$ , $λ x - 3 y = μ$ has infinitely many solutions, then the distance of the point $(λ, μ, - \frac{1}{2})$ from the plane $8 x + y + 4 z + 2 = 0$ is :

Find all values of $λ$ , such that $x, y, z \neq = (0, 0, 0)$ and $(\hat{i} + \hat{j} + 3 \hat{k}) x + (3 \hat{i} - 3 \hat{j} + \hat{k}) y + (- 4 \hat{i} + 5 \hat{j}) z = λ (x \hat{i} + y \hat{j} + z \hat{k})$ where $\hat{i}, \hat{j}, \hat{k}$ are unit vectors along the coordinate axes.

If the system of equations $2 x + 3 y - z = 5$ , $x + α y + 3 z = - 4$ , $3 x - y + β z = 7$ has infinitely many solutions, then $13 α β$ is equal to

Let $S$ be the set of all values of $θ \in [- π, π]$ for which the system of linear equations $x + y + 3 z = 0$ , $- x + (tan θ) y + 7 z = 0$ , $x + y + (tan θ) z = 0$ has non-trivial solution. Then $120 \sum_{θ \in S} θ$ is equal to

If the system of equations $x + (2 sin α) y + (2 cos α) z = 0$ , $x + (cos α) y + (sin α) z = 0$ , $x + (sin α) y - (cos α) z = 0$ has a non-trivial solution, then $α \in (0, \frac{π}{2})$ is equal to :

If the following system of linear equations $2 x + y + z = 5$ , $x - y + z = 3$ , $x + y + a z = b$ has no solution, then :

If the system of linear equations $x + y + 3 z = 0 x + 3 y + k^{2} z = 0 3 x + y + 3 z = 0$ has a non-zero solution $(x, y, z)$ for some $k \in R$ then $x + (\frac{y}{z})^{2}$ is equal to

Select Answer:

Visualized Solution

Setting up the Matrix Equation

First Linear Equation

Second Linear Equation

Third Linear Equation

Solving the System

The Plane Equation

Substituting into Plane Equation

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Evaluating the Expression

Final Result

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Matrix Setup

The Power of Simplification

The Geometric Constraint

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for some α and β in R, the intersection of the following three planes x+4y−2z=1, x+7y−5z=β, x+5y+αz=5 is a line in R3, then α+β is equal to:

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the point P(a,b,c), with reference to (E), lies on the plane 2x+y+z=1, then the value of 7a+b+c is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ω be a solution of x3−1=0 with Im(ω)>0, if a=2 with b and c satisfying (E), then the value of ωa3​+ωb1​+ωc3​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let b=6, with a and c satisfying (E). If α and β are the roots of the quadratic equation ax2+bx+c=0, then ∑n=0∞​(α1​+β1​)n is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of equations 2x+3y−z=5, x+αy+3z=−4, 3x−y+βz=7 has infinitely many solutions, then 13αβ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of equations x+(2​sinα)y+(2​cosα)z=0, x+(cosα)y+(sinα)z=0, x+(sinα)y−(cosα)z=0 has a non-trivial solution, then α∈(0,2π​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the following system of linear equations 2x+y+z=5, x−y+z=3, x+y+az=b has no solution, then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of linear equations x+y+3z=0x+3y+k2z=03x+y+3z=0 has a non-zero solution (x,y,z) for some k∈R then x+(zy​)2 is equal to

Calculating $α$

Coordinates of Point $P$

If for some $α$ and $β$ in $R$ , the intersection of the following three planes $x + 4 y - 2 z = 1$ , $x + 7 y - 5 z = β$ , $x + 5 y + α z = 5$ is a line in $R^{3}$ , then $α + β$ is equal to:

If the point $P (a, b, c)$ , with reference to (E), lies on the plane $2 x + y + z = 1$ , then the value of $7 a + b + c$ is

Let $ω$ be a solution of $x^{3} - 1 = 0$ with $Im (ω) > 0$ , if $a = 2$ with $b$ and $c$ satisfying (E), then the value of $\frac{3}{ω ^{a}} + \frac{1}{ω ^{b}} + \frac{3}{ω ^{c}}$ is equal to

Let $b = 6$ , with $a$ and $c$ satisfying (E). If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ , then $\sum_{n = 0}^{\infty} (\frac{1}{α} + \frac{1}{β})^{n}$ is

If the system of equations $2 x + 3 y - z = 5$ , $x + α y + 3 z = - 4$ , $3 x - y + β z = 7$ has infinitely many solutions, then $13 α β$ is equal to

If the system of equations $x + (2 sin α) y + (2 cos α) z = 0$ , $x + (cos α) y + (sin α) z = 0$ , $x + (sin α) y - (cos α) z = 0$ has a non-trivial solution, then $α \in (0, \frac{π}{2})$ is equal to :

If the following system of linear equations $2 x + y + z = 5$ , $x - y + z = 3$ , $x + y + a z = b$ has no solution, then :

If the system of linear equations $x + y + 3 z = 0 x + 3 y + k^{2} z = 0 3 x + y + 3 z = 0$ has a non-zero solution $(x, y, z)$ for some $k \in R$ then $x + (\frac{y}{z})^{2}$ is equal to