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Consider the line $L_{1} : x = y = z$ and the line $L_{2} : 2 x + y + z - 1 = 0 = 3 x + y + 2 z - 2$ then:

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perpendicular distance of origin from plane containing line $L_{2}$ and parallel to line $L_{1}$ is $\frac{1}{\sqrt{2}}$

the shortest distance between the two lines is $\frac{1}{\sqrt{2}}$

Plane containing the line and parallel to line $L_{1}$ is $z - x + 1 = 0$

The shortest distance between the two lines is $\sqrt{2}$

answer is A, D.

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Detailed Solution

$L_{1} : x = y = z$

$L_{2} : \frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{x}{- 1}$

Shortest distance = $\frac{1}{\sqrt{2}}$

Equation of plane containing line $L_{2}$ and parallel to $L_{1}$

Distance of origin from this plane $= \frac{1}{\sqrt{2}}$

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