The distance of the point (3, 8, 2) from the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$ measured parallel to the plane $3x + 2y - 2z + 15 = 0, $ is

7 

Let P(3, 8, 2) be the given point.

Let the equation of a line through P(3, 8, 2) and parallel to the plane 3x + 2y - 2z + 15 = 0 be

$\frac{x-3}{a}=\frac{y-8}{b}=\frac{z-2}{c}$ ..............(i)

Then, 3a + 2b - 2c = 0 .......(ii)

Line (i) intersects the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$ at Q.

$∴ \begin{vmatrix}3-1 & 8-3 & 2-2\\a & b & c\\2 & 4 & 3\end{vmatrix}=0 ⇒ 15a-6b-2c=0$ .......(iii)

Solving (ii) and (iii) by cross-multiplication, we get

$\frac{a}{-16}=\frac{b}{-24}=\frac{c}{-48}$ or $\frac{a}{2}=\frac{b}{3}=\frac{c}{6}$

Substituting a, b, c in (i), we get

$\frac{x-3}{2}=\frac{y-8}{3}=\frac{z-2}{6}$ as the equation line PQ.

Let the coordinates of Q be (2λ + 3, 3λ + 8, 6λ + 2).

It lies on the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$

$∴ λ + 1 = \frac{3λ+5}{4}= 2λ ⇒ λ = 1 $

Hence, $PQ = \sqrt{(5-3)^2+(11-8)^2+(8-2)^2}= 7 $