The shortest distance b/w the line $\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$ and $x+y+z= 0, 2x-y+z+3= 0$ is   Option: 1 Option: 2 Option: 3 Option: 4

The shortest distance between the lines

$\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1} \text { and } x+y+z+1=0=2x-y+z+3=0$

\begin{aligned} &\text { Line of intersection of planes } \mathrm{x}+\mathrm{y}+\mathrm{z}+1=0=2 \mathrm{x}-\mathrm{y}+\mathrm{z}+3 \text { is }\\ &\text { Eliminating y gives } 3 x+2 z+4=0 \end{aligned}

$\Rightarrow \mathrm{x}=\frac{-2 \mathrm{z}-4}{3}$

\begin{aligned} &\text { Substituting above } \mathrm{x} \text { in } \mathrm{x}+\mathrm{y}+\mathrm{z}+1=0 \text { gives } 3 \mathrm{y}+\mathrm{z}-1=0\\ &\Rightarrow \mathrm{z}=-3 \mathrm{y}+1-(2) \end{aligned}

$\text { From (1) and (2), line equation is } \frac{3 x+4}{-2}=-3 y+1=z$

$\Rightarrow \frac{\mathrm{x}-(-4 / 3)}{-2 / 3}=\frac{\mathrm{y}-(1 / 3)}{-1 / 3}=\frac{\mathrm{z}}{1}$

Given line is  $\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$

Shortest distance between above two skew lines is  $\frac{(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}})}{|\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}|}$

$\begin{array}{l} \overrightarrow{\mathrm{a}}=(1,-1,0) \\\\ \overrightarrow{\mathrm{b}}=\left(-\frac{4}{3}, \frac{1}{3}, 0\right) \\\\ \overrightarrow{\mathrm{c}}=(0,-1,1) \\\\ \overrightarrow{\mathrm{d}}=\left(-\frac{2}{3},-\frac{1}{3}, 1\right) \end{array}$

$\therefore \frac{(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}})}{|\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}|}=\left|\begin{array}{|ccc|} 7 / 3 & -4 / 3 & 0 \\ 0 & -1 & 1 \\ -2 / 3 & -1 / 3 & 1 \\ \hline \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 0 & -1 & 1 \\ -2 / 3 & -1 / 3 & 1 \end{array}\right|=\frac{1}{\sqrt{3}}$

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