The distance of the point (- 1, 2, 3) from the plane → r · ( ^ i -...

The direction ratio of line parallel to the line of the shortest distance between the lines\overset{\to}{r} = ( \overset{^}{i} - \overset{^}{j} ) + \lambda ( 2\overset{^}{i} + \overset{^}{k} ) and \overset{\to}{r} = ( 2\overset{^}{i} - \overset{^}{j} ) + \mu ( \overset{^}{i} - \overset{^}{j} + \overset{^}{k} ) is given by,</p><p>\left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 2 & 0 & 1 \\ 1 & - 1 & 1\end{cases}\right| = \overset{^}{i} - \overset{^}{j} - 2\overset{^}{k}$</p><p>Now given line passing through A ( – 1, 2, 3 )$

And having direction ratio $( 1, – 1, – 2 )$,

So, the equation is given by, $</p><p>\dfrac{x + 1}{1} = \dfrac{y - 2}{- 1} = \dfrac{z - 3}{- 2} =$\lambda$</p><p>Let the line intersect the plane at$P$</p><p>$P ≡ ( \lambda - 1, - \lambda + 2, - 2\lambda + 3 )$</p><p>$P$lies on the plane$x – 2y + 3z = 10$</p><p>$∴ ( \lambda - 1 ) - 2 ( - \lambda + 2 ) + 3 ( - 2\lambda + 3 ) = 10$</p><p>$⇒ \lambda = - 2$</p><p>Hence, point$P ≡ \left(- 3, 4, 7\right)$</p><p>So, the distance of point$P\left(- 3, 4, 7\right) from point A \left(- 1, 2, 3\right) will be, $</p><p>$AP = \sqrt{4 + 4 + 16} = 2\sqrt{6}