Consider the lines $x = y = z$ and the line $2 x + y + z - 1 = 0 = 3 x + y + 2 z - 2$ is

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The shortest distance between the two lines $\frac{\sqrt{3}}{2}$

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

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

Plane containing 2nd line parallel to 1st line is $y - z + 1 = 0$

answer is A, C.

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Detailed Solution

Any plane through the second line is $2 x + y + z - 1 + k (3 x + y + 2 z - 2) = 0$
If this is parallel to the first line, then
$\begin{array}{l} (2 + 3 k) + (1 + k) + (1 + 2 k) = 0 \Rightarrow K = - \frac{2}{3} \\ \Rightarrow Plane is 2 x + y + z - 1 - \frac{2}{3} (3 x + y + 2 z - 2) = 0 \end{array}$
Or $y - z + 1 = 0.$ The required SD must be distance of this plane form any point on the
Line $x = y = z$ say (1, 1, 1)
$\Rightarrow S D = \frac{| 1 - 1 + 1 |}{\sqrt{0^{2} + 1^{2} + {(- 1)}^{2}}} = \frac{1}{\sqrt{2}}$