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

$(2, 6, -2)$

The correct answer is Option (1) → $(2, 6, -2)$ ##

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

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

Now, the co-ordinates of any points $Q$ on the line are $(-2\lambda + 4, 6\lambda, -3\lambda + 1)$.

Also, the given point is $P(2, 3, -8)$

The direction ratios of $PQ$ are $-2\lambda + 4 - 2, 6\lambda - 3, -3\lambda + 1 + 8$

i.e., $-2\lambda + 2, 6\lambda - 3, -3\lambda + 9$

Also, the direction cosines of the given line are $-2, 6, -3$.

If $PQ \perp$ line, then

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

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

$49\lambda - 49 = 0$

$\lambda = 1$

Now, the foot of the perpendicular is $[-2(1) + 4, 6(1) + 1]$ i.e., $(2, 6, -2)$

Hence, the distance $PQ$ is

$= \sqrt{(2-2)^2 + (3-6)^2 + (-8+2)^2}$

$= \sqrt{0 + 9 + 36} = \sqrt{45} = 3\sqrt{5} \text{ units}$