The distance of the point (-1, 2, 6) from the line $\frac{x-2}{6}=\frac{y-3}{3}=\frac{z+4}{-4}$, is equal to :

7 units

Any point on the line is

P = (6r₁ + 2, 3r₁ + 3, –4r₁ – 4)

Direction ration of the line segment PQ, where Q = (–1, 2, 6), are

6r₁ + 3, 3r₁ + 1, – 4r, – 10

If ‘P’ be the foot of altitude drawn from Q to the given line, then

6(6r₁ + 3) + 3(3r₁ + 1) + 4(4r₁ + 10) = 0

⇒ r₁ = –1

Thus, P = (–4, 0, 0)

∴ Required distance = $\sqrt{9+4+36}$

= 7 units