. Find the foot of perpendicular from the point (2,3,-8) to the line $\frac{{4 - x}}{2} = \frac{y}{6} = \frac{{1 - z}}{3}$. Also, find the perpendicular distance from the given point to the line.

It is given that, equation of line as $\frac{{4 - x}}{2} = \frac{y}{6} = \frac{{1 - z}}{3}$

$\Rightarrow$ $\frac{{x - 4}}{{ - 2}} = \frac{y}{6} = \frac{{z - 1}}{{ - 3}} = \lambda$

$\Rightarrow$ $x = - 2\lambda + 4,y = 6\lambda$ and $z = - 3\lambda + 1$

Let the coordinates of $L$ be $(4 - 2\lambda ,6\lambda ,1 - 3\lambda )$ and direction ratios of PL are proportional to

$(4 - 2\lambda - 2,6\lambda - 3,1 - 3\lambda + 8)$ i.e., $(2 - 2\lambda ,6\lambda - 3,9 - 3\lambda )$

Also, direction ratios are proportional to -2,6,-3 . Since, PL is perpendicular to give line.

$\therefore - 2(2 - 2\lambda ) + 6(6\lambda - 3) - 3(9 - 3\lambda ) = 0$

$\Rightarrow$ $- 4 + 4\lambda + 36\lambda - 18 - 27 + 9\lambda = 0$

$\Rightarrow$ $49\lambda = 49 \Rightarrow \lambda = 1$

So, the coordinates of $L$ are $(4 - 2\lambda ,6\lambda ,1 - 3\lambda )$ i.e., (2,6,-2) .

$\frac{{4 - x}}{2} = \frac{y}{6} = \frac{{1 - z}}{3}$

Also, length of $PL = \sqrt {{{(2 - 2)}^2} + {{(6 - 3)}^2} + {{( - 2 + 8)}^2}}$
$= \sqrt {0 + 9 + 36} = 3\sqrt 5$ units