Let P be the plane passing through the point (1, 2, 3) and the line of intersection of the planes r

Q: Why do we use $\lambda$ instead of just solving for $x, y, z$ directly?

Great question! The line of intersection contains an infinite number of points. We cannot solve for a unique point because the system is underdetermined. However, we have enough information to define the *entire plane* by using the parameter $\lambda$, which acts as a 'tuning knob' to select the specific plane that passes through our target point $(1, 2, 3)$.

Q: Is there a risk of losing a plane in this method?

Technically, yes. The equation $P_1 + \lambda P_2 = 0$ covers all planes passing through the intersection *except* for $P_2$ itself. In a rigorous JEE Advanced scenario, always keep this in mind. However, for this problem, the point $(1, 2, 3)$ does not lie on $P_2$, so we are safe to use the $\lambda$ method without worry.

Q: Why did the question ask for the point that does NOT lie on the plane?

This is a classic JEE Advanced tactic to test your precision and attention to detail. Many students find the equation of the plane and then immediately mark the first option that satisfies it. By asking for the point that does *not* lie on the plane, the examiner forces you to verify every single option. It is a test of your patience and your ability to avoid 'autopilot' mode.

Question

Let $P$ be the plane passing through the point $(1, 2, 3)$ and the line of intersection of the planes $\vec{r} \cdot (\hat{i} + \hat{j} + 4\hat{k}) = 16$ and $\vec{r} \cdot (-\hat{i} + \hat{j} + \hat{k}) = 6$. Then which of the following points does NOT lie on $P$?

Options:

(A) $(3, 3, 2)$

(B) $(6, -6, 2)$

(C) $(4, 2, 2)$

(D) $(-8, 8, 6)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a coordinate geometry problem; we are learning to visualize the architecture of 3D space.

Imagine you are holding an open book. The spine of the book is a line, and two pages meet at that spine. These are our two given planes:
$$P_1: x + y + 4z - 16 = 0$$
$$P_2: -x + y + z - 6 = 0$$

The problem asks us to find a third plane, $P$, that passes through this spine. This is the essence of the **Family of Planes** concept. Think of it as a door hinge; we are looking for the specific angle of the door that allows it to pass through a specific point in space, $(1, 2, 3)$.

### Translating the Language of Vectors

Before we can dance with the geometry, we must speak the language. The problem gives us planes in vector form:
$$\vec{r} \cdot (\hat{i} + \hat{j} + 4\hat{k}) = 16$$
$$\vec{r} \cdot (-\hat{i} + \hat{j} + \hat{k}) = 6$$

To make this tangible, we replace the position vector $\vec{r}$ with $x\hat{i} + y\hat{j} + z\hat{k}$. This substitution transforms the abstract vector notation into the familiar Cartesian form:
$$x + y + 4z - 16 = 0$$
$$-x + y + z - 6 = 0$$

We have now grounded our problem in the $xyz$-coordinate system.

### The Power of the Family

Now, we invoke the most elegant tool in our arsenal: the **Family of Planes** equation. Any plane passing through the intersection of $P_1$ and $P_2$ can be expressed as $P_1 + \lambda P_2 = 0$.

This works because it is a linear combination that satisfies both plane equations simultaneously at the line of intersection. Substituting our planes, we get:
$$(x + y + 4z - 16) + \lambda(-x + y + z - 6) = 0$$

This equation represents every possible plane that could pass through that spine. We need the specific one that passes through $(1, 2, 3)$.

### Locking the Plane

This is where we apply our constraint. If the plane passes through $(1, 2, 3)$, then these coordinates must satisfy our equation. Substituting $x=1, y=2, z=3$ into our family equation:
$$(1 + 2 + 4(3) - 16) + \lambda(-1 + 2 + 3 - 6) = 0$$

Let us calculate carefully:
The first bracket yields $1 + 2 + 12 - 16 = -1$.
The second bracket yields $-1 + 2 + 3 - 6 = -2$.

Thus, we are left with $-1 - 2\lambda = 0$, which gives us $\lambda = -\frac{1}{2}$.

By substituting $\lambda = -\frac{1}{2}$ back into our family equation, we define our unique plane $P$:
$$2(x + y + 4z - 16) - 1(-x + y + z - 6) = 0$$

Expanding this, we arrive at:
$$2x + 2y + 8z - 32 + x - y - z + 6 = 0$$

Simplifying, we get the final equation of the plane:
**$$3x + y + 7z - 26 = 0$$**

### The Final Verification

The question asks for the point that does NOT lie on this plane. This is a trap for the hasty. We must test the given options against our derived equation.

Let us test the point $(4, 2, 2)$:
$$3(4) + 2 + 7(2) - 26 = 12 + 2 + 14 - 26 = 28 - 26 = 2$$

Since $2 
eq 0$, this point does not satisfy the equation and therefore does not lie on the plane. We have successfully navigated the trap.

Remember, in JEE Advanced, the math is the path, but the logic is the destination. Keep practicing, keep visualizing, and stay curious.

Select Answer:

Visualized Solution

Convert Vector to Cartesian Form

Family of Planes Concept

Substitute the Given Point

Simplify the Brackets

Find the Value of $λ$

Substitute $λ$ into Plane Equation

Multiply by $2$

Simplify to Final Equation

Test Option 3: $(4, 2, 2)$

Final Verification

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Translating the Language of Vectors

The Power of the Family

Locking the Plane

The Final Verification

Similar Questions

Let the lines $\frac{x - 1}{λ} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x + 26}{- 2} = \frac{y + 18}{3} = \frac{z + 28}{λ}$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does NOT lies on $P$ ?

A plane $P$ contains the line $x + 2 y + 3 z + 1 = 0 = x - y - z - 6$ , and is perpendicular to the plane $- 2 x + y + z + 8 = 0$ . Then which of the following points lies on $P$ ?

Let $Q$ be the mirror image of the point $P (1, 2, 1)$ with respect to the plane $x + 2 y + 2 z = 16$ . Let $T$ be a plane passing through the point $Q$ and contains the line $r = - \hat{k} + λ (\hat{i} + \hat{j} + 2 \hat{k}), λ \in R$ . Then, which of the following points lies on $T$ ?

The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2 x + 3 y - z = 5$ , contains which one of the following points ?

The plane passing through the point (4, -1, 2) and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point :

The plane which bisects the line segment joining the points $(- 3, - 3, 4)$ and $(3, 7, 6)$ at right angles, passes through which one of the following points ?

The plane passing through the points $(1, 2, 1)$ , $(2, 1, 2)$ and parallel to the line, $2 x = 3 y, z = 1$ also passes through the point: