The shortest distance between the pair of skew lines:
$x = - y + 2 = - z + 2$ and $x - 2 = - y + 1 = z + 1$ is $\frac{1}{\sqrt{n}}$ . Then the value of $n$ is:

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Solution

Given lines :
$L_{1} : x = - y + 2 = - z + 2$ and $L_{2} : x - 2 = - y + 1 = z + 1$
$\Rightarrow L_{1} : \frac{x}{1} = \frac{y - 2}{- 1} = \frac{z - 2}{- 1}, \Rightarrow L_{2} : \frac{x - 2}{1} = \frac{y - 1}{- 1} = \frac{z + 1}{1}$
$D . R^{'} s$ of $L_{1} = (a_{1}, b_{1}, c_{1}) = (1, - 1, - 1)$ and $D . R^{'} s$ of $L_{2} = (a_{2}, b_{2}, c_{2}) = (1, - 1, 1)$
$∴$ shortest distance $= ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \frac{∣ ∣ ∣ ∣ \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} \end{matrix} ∣ ∣ ∣ ∣}{\sqrt{\sum (b_{1} c_{2} - b_{2} c_{1})^{2}}} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \frac{∣ ∣ ∣ ∣ \begin{matrix} 2 & - 1 & - 3 1 & - 1 & - 1 1 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣}{\sqrt{(- 1 - 1)^{2} + (- 1 - 1)^{2} + (- 1 + 1)^{2}}} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = \frac{2}{2 \sqrt{2}} = \frac{1}{\sqrt{2}}$
So, $n = 2$

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