Let P be the plane containing the straight line (x-3)/9 = (y+4)/-1 = (z-7)/-5 and perpendicular to t

Q: Why do we cross $\vec{v}_2$ and $\vec{v}_3$ to get the normal of $Q$?

Great question! Remember that a plane is defined by its normal vector. If two lines lie on a plane, their direction vectors also lie on that plane. The cross product of any two non-parallel vectors lying on a plane will always yield a vector perpendicular to that plane—which is exactly our normal vector $\vec{n}_Q$.

Q: Can I pick any point on line $L_1$ to write the equation of plane $P$?

Absolutely! The equation of a plane is determined by its normal vector and *any* point that lies on it. Since the entire line $L_1$ is contained within plane $P$, every single point on that line is a valid point for your equation. You will get the same plane equation regardless of which point you choose.

Q: Why is the distance formula $d^2$ and not $d$?

The question specifically asks for $d^2$. This is a classic JEE trap to test if you are reading the question carefully. Always calculate the distance $d$ first, and then square it at the very final step. Don't let a simple calculation error at the end ruin your hard work!

Question

Let $P$ be the plane containing the straight line $\frac{x-3}{9} = \frac{y+4}{-1} = \frac{z-7}{-5}$ and perpendicular to the plane containing the straight lines $\frac{x}{2} = \frac{y}{3} = \frac{z}{5}$ and $\frac{x}{3} = \frac{y}{7} = \frac{z}{8}$. If $d$ is the distance of $P$ from the point $(2, -5, 11)$, then $d^2$ is equal to :

Options:

(A) $\frac{147}{2}$

(B) $96$

(C) $\frac{32}{3}$

(D) $54$

Sigma Percentile · Accepted Answer

### The Architecture of 3D Space: A Journey Through Planes and Normals

Welcome, future engineer. Today, we are not just solving a problem; we are constructing a 3D architecture. We are dealing with planes, lines, and the beautiful, rigid logic that binds them together.

When you look at a problem involving planes, don't just see equations. See the geometry. Imagine a sheet of paper (Plane $P$) standing upright, intersecting another sheet (Plane $Q$) at a perfect right angle. This is the world we are about to map.

### Phase 1: Decoding the Foundation of Plane $Q$

Our journey begins with Plane $Q$. We are told it contains two lines, $L_2$ and $L_3$. In the language of vectors, if a plane contains a line, it contains the line's direction vector.

So, we extract the direction vectors $\vec{v}_2 = (2, 3, 5)$ and $\vec{v}_3 = (3, 7, 8)$. These vectors are the DNA of Plane $Q$. To find the normal vector $\vec{n}_Q$, we need a vector that is perpendicular to both.

The cross product is our tool of choice: $\vec{n}_Q = \vec{v}_2 	imes \vec{v}_3$. Calculating this determinant:

$$
\vec{n}_Q = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 5 \ 3 & 7 & 8 \end{vmatrix} = -11\hat{i} - \hat{j} + 5\hat{k}
$$

This vector is the anchor of Plane $Q$.

### Phase 2: The Dance of Normals for Plane $P$

Now, we pivot to Plane $P$. We know it contains line $L_1$, which gives us the direction vector $\vec{v}_1 = (9, -1, -5)$ and a point $A = (3, -4, 7)$.

But here is the crux: Plane $P$ is perpendicular to Plane $Q$. This implies their normal vectors are also perpendicular. So, $\vec{n}_P$ must be perpendicular to $\vec{n}_Q$.

Furthermore, since $L_1$ lies in $P$, $\vec{n}_P$ must be perpendicular to $\vec{v}_1$. We have two vectors, $\vec{v}_1$ and $\vec{n}_Q$, and we need a vector perpendicular to both. Again, the cross product saves us: $\vec{n}_P = \vec{v}_1 	imes \vec{n}_Q$.

$$
\vec{n}_P = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 9 & -1 & -5 \ -11 & -1 & 5 \end{vmatrix} = -10\hat{i} + 10\hat{j} - 20\hat{k}
$$

### Phase 3: Constructing the Plane and the Final Leap

We can simplify $\vec{n}_P$ by dividing by $-10$, giving us the direction ratios $(1, -1, 2)$. With this normal and our point $A(3, -4, 7)$, we write the equation:

$$
1(x-3) - 1(y+4) + 2(z-7) = 0
$$

Simplifying this, we arrive at the elegant equation of Plane $P$: $x - y + 2z - 21 = 0$.

Finally, we calculate the distance $d$ from the point $(2, -5, 11)$ using the formula:

$$
d = \frac{|ax_0 + by_0 + cz_0 + D|}{\sqrt{a^2 + b^2 + c^2}}
$$

Substituting our values, we get:

$$
d = \frac{|1(2) - 1(-5) + 2(11) - 21|}{\sqrt{1^2 + (-1)^2 + 2^2}} = \frac{8}{\sqrt{6}}
$$

Squaring this gives us the final result:

$$
d^2 = \frac{64}{6} = \mathbf{\frac{32}{3}}
$$

You have successfully navigated the geometry. Take a moment to appreciate the symmetry—the way the cross products acted as the bridge between the lines and the planes. This is the elegance of JEE mathematics.

Select Answer:

Visualized Solution

Visualizing the Planes

Direction Vectors of $L_{2}$ and $L_{3}$

Finding Normal to Plane $Q$

Calculating $n_{Q}$

Line $L_{1}$ in Plane $P$

Normal to Plane $P$

Setting up $n_{P}$ Cross Product

Calculating $n_{P}$

Simplifying Normal $P$

Equation of Plane $P$ Setup

Final Equation of Plane $P$

Point $M$ and Distance $d$

Calculating $d$

Evaluating $d$

Final Result: $d^{2}$

The Sigma Insight: Equation of a Plane

The Architecture of 3D Space

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Visualizing the Planes

Direction Vectors of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L3​

Finding Normal to Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

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

Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ in Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Normal to Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nP​ Cross Product

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

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

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

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

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M and Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

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

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

Final Result: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2