If a line passing through (2, 1, 4) cuts off an intercept of minimum length between two non coplanar lines $x - 6 = \frac{y}{α} = - z$
and $x = \frac{y}{0} = z$ then $α$ is

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-2

$\frac{4}{3}$

$\frac{3}{2}$

answer is D.

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Detailed Solution

Line passes through $P (2, 1, 4)$ lie along shortest distance between two non – coplanar lines

$L_{1} = \frac{x - 6}{1} = \frac{y}{α} = \frac{z}{- 1} = k_{1} \Rightarrow A (6 + k_{1}, α k_{1}, - k_{1})$

$L_{2} = \frac{x}{1} = \frac{y}{0} = \frac{z}{1} = k_{2} \Rightarrow B (k_{2}, 0, k_{2})$

$\begin{array}{l} {PA perpendicular to L}_{1} \Rightarrow (6 + k_{1} - 2) 1 + (α k_{1} - 1) α + (- k_{1} - 4) (- 1) = 0 \\ α^{2} k_{1} - α + 2 k_{1} + 8 = 0 \\ P B {is perpendicular to L}_{2} \Rightarrow k_{2} = 3 \\ A B perpendicualr to L_{1} \\ (6 + k_{1} - 3) + α (α k_{1}) + (- 1) (- k_{1} - 3) = 0 \\ α^{2} k_{1} + 2 k_{1} + 6 = 0 \\ Hence, α =2 \end{array}$