The equation of the plane passing through the line of intersection of the planes $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$ and parallel to the $x-$ axis is

(C) $\vec{r} \cdot(\hat{j}-3 \hat{k})+6=0$

(A) $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{j}}-3 \hat{\mathrm{k}})-6=0$

(B) $\vec{r} \cdot(\hat{i}+3 \hat{k})+6=0$

(D) $\vec{r} \cdot(\hat{i}-3 \hat{k})+6=0$

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Three Dimensional Geometry

The equation of the plane passing through the line of intersection of the planes

r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1

and

r \cdot (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0

and parallel to the

x -

axis is

Answer: (C)

The family of planes through the line of intersection is

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

, i.e.,

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

. For the plane to be parallel to the

x

-axis, the normal must be perpendicular to

\hat{i}

, so the coefficient of

x

is zero:

1 + 2 λ = 0

gives

λ = - 1/2

. Substituting:

0 \cdot x + (1 - 3/2) y + (1 + 1/2) z + (- 2 - 1) = 0

, i.e.,

- \frac{1}{2} y + \frac{3}{2} z - 3 = 0

, or

y - 3 z + 6 = 0

, which is

r \cdot (\hat{j} - 3 \hat{k}) + 6 = 0

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option