If the lines $\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{j}}-\hat{\mathrm{k}})$ and $\overrightarrow{\mathrm{r}}=(\alpha \hat{\mathrm{i}}-\hat{\mathrm{j}})+\mu(2 \hat{\mathrm{i}}-3 \hat{\mathrm{k}})$ are co-planar, then distance of the plane containing these two lines from the point $(\alpha, 0,0)$ is :

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Three Dimensional Geometry

If the lines

r = (\hat{i} - \hat{j} + \hat{k}) + λ (3 \hat{j} - \hat{k})

and

r = (α \hat{i} - \hat{j}) + μ (2 \hat{i} - 3 \hat{k})

are co-planar, then distance of the plane containing these two lines from the point

(α, 0, 0)

is :

Answer: (B)

For coplanarity, the determinant

02 1 - α 300 - 1 - 3 1 = 0

.

Expanding:

- 3 (2 + 3 (1 - α)) = 0

, which gives

2 + 3 - 3 α = 0

, so

α = \frac{5}{3}

.

The normal vector is

n = \hat{i} 02 \hat{j} 30 \hat{k} - 1 - 3 = - 9 \hat{i} - 2 \hat{j} - 6 \hat{k}

.

The plane equation through

(1, - 1, 1)

9 x + 2 y + 6 z - 13 = 0

.

Distance from

(\frac{5}{3}, 0, 0)

\frac{∣9 \cdot \frac{5}{3} - 13∣}{81 + 4 + 36} = \frac{2}{11}

.

Answer: (B)

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Single correct option