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

Let $P(3,8,2)$ be the given point and $L: \frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3}$ the line. The parametric form of $L$ gives $Q(1 + 2t, 3 + 4t, 2 + 3t)$ for some $t \in \mathbb{R}$.

The plane $\Pi: 3x + 2y - 2z + 15 = 0$ has normal vector $\vec{n} = (3, 2, -2)$.
The distance from $P$ to $L$ measured parallel to $\Pi$ is the length of the segment $PQ$ where $\vec{PQ}$ is parallel to $\Pi$, meaning $\vec{PQ} \cdot \vec{n} = 0$.

The vector $\vec{PQ} = Q - P = (2t - 2, 4t - 5, 3t)$.
Setting the dot product to zero: $(2t - 2)(3) + (4t - 5)(2) + (3t)(-2) = 0$
Simplifying: $6t - 6 + 8t - 10 - 6t = 0$
$8t - 16 = 0$
$t = 2$

Substituting $t = 2$ into $Q$ yields $Q(5, 11, 8)$.
The distance is $|PQ| = \sqrt{(5 - 3)^2 + (11 - 8)^2 + (8 - 2)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

Final answer: $7$ units