The line l _ 1 passes through the point (2, 6, 2) and is...

The line ${l}_{1}$ passes through the point $( 2, 6, 2 )$ and is perpendicular to the line $2x + y - 2z = 10$ then,

${l}_{1}: \dfrac{x - 2}{2} = \dfrac{y - 6}{1} = \dfrac{z - 2}{- 2}$

Let the shortest distance between ${l}_{1}$ and the line $\dfrac{x + 1}{2} = \dfrac{y + 4}{- 3} = \dfrac{z}{2}$ is S.D.

S.D. $= \left|\dfrac{\overset{\to}{AB} \cdot \overset{\to}{MN}}{\left|\overset{\to}{MN}\right|}\right|$

Here, \overset{\to}{AB} = 3\overset{^}{i} + 10\overset{^}{j} + 2\overset{^}{k}

\overset{\to}{MN} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 2 & 1 & - 2 \\ 2 & - 3 & 2\end{cases}\right| = - 4\overset{^}{i} - 8\overset{^}{j} - 8\overset{^}{k}

$⇒ \left|\overset{\to}{MN}\right| = \sqrt{16 + 64 + 64} = 12$

$\overset{\to}{AB} \cdot \overset{\to}{MN} = - 12 - 80 - 16 = - 108$

So, shortest distance (S.D.)

$= \left|\dfrac{- 108}{12}\right| = \left|- 9\right| = 9$ units