Explanation for the correct option:
Given: line $\frac{\mathit{x}-2}{3}=\frac{\mathit{y}-3}{4}=\frac{\mathit{z}-4}{5}$
let $\frac{\mathit{x}-2}{3}=\frac{\mathit{y}-3}{4}=\frac{\mathit{z}-4}{5}=\mathit{\lambda }$
$3\hat{\mathit{i}}+4\hat{\mathit{j}}+5\hat{\mathit{k}}=\overrightarrow{\mathit{a}}$ will be perpendicular to normal vector of a plane.
Checking option $\mathit{D}$
So,
$=3\left(2\right)+4\left(1\right)+5\left(-2\right)$
$=6+4-10$
$=0$
Hence, $2\mathit{x}+\mathit{y}-2\mathit{z}=0$ is parallel to $\frac{\mathit{x}-2}{3}=\frac{\mathit{y}-3}{4}=\frac{\mathit{z}-4}{5}$.