Let α, β and γ be real numbers such that the system of linear equations
x + 2y + 3z = α
4x + 5y + 6

Q: Wait, why is the determinant of the coefficient matrix zero? Does that mean there is no solution?

Not necessarily! A zero determinant means the system is either inconsistent (no solution) or dependent (infinitely many solutions). The problem explicitly states the system is consistent, which forces us into the 'infinitely many solutions' scenario. It is a beautiful hint that the equations are not independent.

Q: How did you know to write $E_3 = A \cdot E_2 + B \cdot E_1$? Is there a faster way?

You could use the augmented matrix and perform row reduction (RREF), but the method of linear combinations is often faster for $3 \times 3$ systems. By comparing coefficients, you are essentially performing the row operations mentally. It is a powerful tool for revealing the hidden structure of the system.

Q: Why does the plane $P$ have the equation $x - 2y + z = 1$? Where did the variables come from?

The consistency condition $\alpha - 2\beta + \gamma = 1$ defines the set of all points $(\alpha, \beta, \gamma)$ that allow the system to exist. If we treat these parameters as coordinates in 3D space, this linear equation describes a flat surface—a plane. We simply swap the labels to $x, y, z$ to visualize it in standard Cartesian form.

Q: I got $d = 3/\sqrt{6}$. Why is the answer $1.5$?

Always read the final line of the question carefully! The problem asks for $D$, which is defined as the *square* of the distance. Squaring $3/\sqrt{6}$ gives $9/6$, which simplifies perfectly to $1.5$. It is a classic JEE trap to make you stop at the distance calculation.

Q: Is it a coincidence that $|M|$ equals the constant in the plane equation?

It is not a coincidence at all—it is the mathematical 'soul' of the problem! The structure of matrix $M$ was specifically designed to mirror the consistency condition. Recognizing this symmetry is what separates a good student from a master.

Question

Passage:
Let $\alpha, \beta$ and $\gamma$ be real numbers such that the system of linear equations 
$x + 2y + 3z = \alpha$
$4x + 5y + 6z = \beta$
$7x + 8y + 9z = \gamma - 1$
is consistent. Let $|M|$ represent the determinant of the matrix $M = \begin{bmatrix} \alpha & 2 & \gamma \ \beta & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$. Let $P$ be the plane containing all those $(\alpha, \beta, \gamma)$ for which the above system of linear equations is consistent, and $D$ be the square of the distance of the point $(0, 1, 0)$ from the plane $P$.

Question 1:
The value of $|M|$ is

Question 2:
The value of $D$ is

Sigma Percentile · Accepted Answer

### Analyzing the Determinant

Imagine you are standing before a system of three linear equations. At first glance, they look like three distinct planes in space, perhaps intersecting at a single point.

But as you calculate the determinant of the coefficient matrix

$$\Delta = \begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{vmatrix}$$

you realize something profound happens. Calculating the expansion:

$$\Delta = 1(45 - 48) - 2(36 - 42) + 3(32 - 35) = -3 + 12 - 9 = 0$$

The determinant vanishes! This is not a failure; it is a revelation. It tells us that these planes are not independent; they are locked in a geometric dance where one is merely a shadow of the others.

### Unmasking the Hidden Relation

Since the system is consistent despite the zero determinant, it must possess infinitely many solutions. This implies that the third equation is a linear combination of the first two.

We seek constants $A$ and $B$ such that the third row is a weighted sum of the first two: $R_3 = A R_2 + B R_1$. By comparing the coefficients of $x, y,$ and $z$, we set up a mini-system:

$$4A + B = 7$$
$$5A + 2B = 8$$

Solving this, we find $A = 2$ and $B = -1$.

This is the key that unlocks the vault. For the system to remain consistent, the constant terms must obey the exact same rule: $C_3 = A C_2 + B C_1$.

Substituting our values, we get $\gamma - 1 = 2(\beta) - 1(\alpha)$, which rearranges into the elegant consistency condition:

$$\alpha - 2\beta + \gamma = 1$$

### The Geometry of Consistency

Now, look at the matrix

$$M = \begin{bmatrix} \alpha & 2 & \gamma \ \beta & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$$

When we expand its determinant $|M|$ along the first row, we get $|M| = \alpha(1) - 2(\beta) + \gamma(1)$.

Look closely—this is exactly the left-hand side of our consistency condition! Because the system is consistent, $|M|$ must equal $1$. This is the beauty of linear algebra; the algebraic constraint on the system is encoded directly into the determinant of $M$.

### The Final Leap: Distance to the Plane

We have identified that all valid points $(\alpha, \beta, \gamma)$ lie on the plane $P$ defined by $x - 2y + z = 1$. We are asked to find the square of the distance $D$ from the point $(0, 1, 0)$ to this plane.

Using the standard distance formula

$$d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$$

we substitute our values:

$$d = \frac{|1(0) - 2(1) + 1(0) - 1|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|-3|}{\sqrt{6}} = \frac{3}{\sqrt{6}}$$

Finally, we calculate the square of the distance:

$$D = d^2 = \left( \frac{3}{\sqrt{6}} ight)^2 = \frac{9}{6} = 1.5$$

Through this journey, we moved from the abstract dependency of rows to the concrete geometry of a plane, finally arriving at the final result of **1.5**. Remember, in JEE Advanced, the math is never just about calculation; it is about seeing the hidden connections between the equations.

Animated Solution for Mathematics - Three Dimensional Geometry: Comprehension Passage

The value of $∣ M ∣$ is

Enter Numerical Value:

The value of $D$ is

Enter Numerical Value:

Visualized Solution

System of Equations

Coefficient Matrix Determinant

Condition for Consistency

Solving for Multipliers

Applying to Constant Terms

Determinant of Matrix M

Evaluating |M|

Defining Plane P

Distance from a Point to a Plane

Calculating Distance d

Calculating D

The Sigma Insight: Equation of a Plane

Analyzing the Determinant

Unmasking the Hidden Relation

The Geometry of Consistency

The Final Leap

Similar Questions

Let the plane $P$ pass through the intersection of the planes $2 x + 3 y - z = 2$ and $x + 2 y + 3 z = 6$ , and be perpendicular to the plane $2 x + y - z + 1 = 0$ . If $d$ is the distance of $P$ from the point $(- 7, 1, 1)$ , then $d^{2}$ is equal to :

If the equation of a plane $P$ , passing through the intersection of the planes $x + 4 y - z + 7 = 0$ and $3 x + y + 5 z = 8$ is $a x + b y + 6 z = 15$ for some $a, b \in R$ , then the distance of the point $(3, 2, - 1)$ from the plane $P$ is

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

Let the plane containing the line of intersection of the planes $P 1 : x + (λ + 4) y + z = 1$ and $P 2 : 2 x + y + z = 2$ pass through the points $(0, 1, 0)$ and $(1, 0, 1)$ . Then the distance of the point $(2 λ, λ, - λ)$ from the plane $P 2$ is

A plane $E$ is perpendicular to the two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , and passes through the point $P (1, - 1, 1)$ . If the distance of the plane $E$ from the point $Q (a, a, 2)$ is $32$ , then $(P Q)^{2}$ is equal to

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

The distance of the point $(1, 3, - 7)$ from the plane passing through the point $(1, - 1, - 1)$ , having normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$ , is:

Let a plane $P$ pass through the point $(3, 7, - 7)$ and contain the line, $\frac{x - 2}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ . If distance of the plane $P$ from the origin is $d$ , then $d^{2}$ is equal to

If the plane $2 x - y + 2 z + 3 = 0$ has the distances $1/3$ and $2/3$ units from the planes $4 x - 2 y + 4 z + λ = 0$ and $2 x - y + 2 z + μ = 0$ , respectively, then the maximum value of $λ + μ$ is equal to :

Animated Solution for Mathematics - Three Dimensional Geometry: Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ∣M∣ is

Enter Numerical Value:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of D is

Enter Numerical Value:

Visualized Solution

System of Equations

Coefficient Matrix Determinant

Condition for Consistency

Solving for Multipliers

Applying to Constant Terms

Determinant of Matrix M

Evaluating |M|

Defining Plane P

Distance from a Point to a Plane

Calculating Distance d

Calculating D

The Sigma Insight: Equation of a Plane

Analyzing the Determinant

Unmasking the Hidden Relation

The Geometry of Consistency

The Final Leap

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane which is perpendicular to two planes 2x−2y+z=0 and x−y+2z=4, passes through (1,−2,1). The distance of the plane from the point (1,2,2) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the distance between the plane Ax−2y+z=d and the plane containing the lines 2x−1​=3y−2​=4z−3​ and 3x−2​=4y−3​=5z−4​ is 6​, then find ∣d∣.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a plane P pass through the point (3,7,−7) and contain the line, −3x−2​=2y−3​=1z+2​. If distance of the plane P from the origin is d, then d2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the plane 2x−y+2z+3=0 has the distances 1/3 and 2/3 units from the planes 4x−2y+4z+λ=0 and 2x−y+2z+μ=0, respectively, then the maximum value of λ+μ is equal to :

The value of $∣ M ∣$ is

The value of $D$ is

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

Let a plane $P$ pass through the point $(3, 7, - 7)$ and contain the line, $\frac{x - 2}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ . If distance of the plane $P$ from the origin is $d$ , then $d^{2}$ is equal to

If the plane $2 x - y + 2 z + 3 = 0$ has the distances $1/3$ and $2/3$ units from the planes $4 x - 2 y + 4 z + λ = 0$ and $2 x - y + 2 z + μ = 0$ , respectively, then the maximum value of $λ + μ$ is equal to :