Consider the lines given as
$L_{1} : (3 \hat{i} - \hat{j} + 2 \hat{k}) + λ (2 \hat{i} - \hat{k}), λ \in R$
$L_{2} : (\hat{i} + 3 \hat{j} - \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k}), μ \in R$
Let P be a plane containing the line $L_{1}$ and parallel to line $L_{2}$ . The plane meets x, y and z axes at points A, B and C respectively. The shortest distance between lines $L_{1}$ and $L_{2}$ equals to D and volume of tetrahedron $OABC$ is equals to V, where O is the origin then

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$\frac{V}{D} = \frac{4 \sqrt{14}}{9}$

$VD = \frac{64}{9 \sqrt{14}}$

$\frac{V}{D} = \frac{8 \sqrt{14}}{9}$

$VD = \frac{8}{9 \sqrt{14}}$

answer is B, C.

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Detailed Solution

The equation of plane is given by $|\begin{matrix} x - 3 & y + 1 & z - 2 \\ 2 & 0 & - 1 \\ 1 & - 1 & 1 \end{matrix}| = 0$

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

$∴ A \equiv (4, 0, 0) \cdot B = (0, 4 / 3, 0) \cdot C = (0, 0, 2)$

$∴ V = \frac{1}{6} | [O \hat{A} O \hat{B} O \hat{C}] \Rightarrow V = \frac{16}{9}$

Now, $D = | \frac{(2 \hat{i} - 4 \hat{j} + 3 \hat{k}) \cdot ((2 \hat{i} - \hat{k}) \times (\hat{i} - \hat{j} + \hat{k})) ∣}{| (2 \hat{1} - \hat{k}) \times (\hat{i} - \hat{j} + \hat{k}) |} |$ $\Rightarrow D = \frac{4}{\sqrt{19}}$

$∴ V D = \frac{64}{9 \sqrt{19}} and \frac{V}{D} = \frac{4 \sqrt{14}}{9}$

$∴$ Option (b), (c) are correct.

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