JEE Mathematics 2023 Question with Solution

Given: The required plane passes through the line of intersection of $P_1: 2x - y + z - 3 = 0$ and $P_2: 4x - 3y + 5z + 9 = 0$, and it is parallel to the line $\frac{x+1}{-2} = \frac{y+3}{4} = \frac{z-2}{5}$.

Find: The value of $a+b+c$ if the plane is $ax + by + cz + 6 = 0$.

A plane through the line of intersection of $P_1$ and $P_2$ is

so,

The given line has direction vector

and the normal vector of the plane is

Since the plane is parallel to the line, $\vec{n} \cdot \vec{v} = 0$.

Therefore,

Substituting $\lambda = \frac{3}{5}$ in the plane equation,

Comparing with $ax + by + cz + 6 = 0$, we get

Hence,

Therefore, the correct option is B.

The family of planes through the intersection of two planes is formed by taking their linear combination. This guarantees that every plane in the family contains the common line of intersection.

Next, use the fact that if a line is parallel to a plane, then the direction ratios of the line are perpendicular to the normal of the plane. That is why the dot product condition

is applied.

Solving this gives $\lambda = \frac{3}{5}$, and substituting into the family equation gives the plane

from which

So the required value is $14$.