Q42 / 90 · JEE Main 2019 (08 Apr Shift 2) · Mathematics

The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is

Solution

Planes passing through line of intersection of the planes

x + y + z = 1

and

2 x + 3 y + 4 z = 5

(x + y + z - 1) + λ (2 x + 3 y + 4 z - 5) = 0

\Rightarrow (1 + 2 λ) x + (1 + 3 λ) y + (1 + 4 λ) z - (1 + 5 λ) = 0

Since required plane is perpendicular to

x - y + z = 0,

hence

1 \cdot (1 + 2 λ) - 1 \cdot (1 + 3 λ) + 1 \cdot (1 + 4 λ) = 0 \Rightarrow λ = - \frac{1}{3}

Hence, required plane is

x - z + 2 = 0

i.e.

\vec{r} \cdot (\hat{i} - \hat{k}) + 2 = 0

Hence, the correct answer is option: 'D'.