The line $\dfrac{x-3}{4}=\dfrac{y-4}{5}=\dfrac{z-5}{6}$ is parallel to the plane

The direction vector of the given line $\dfrac{x-3}{4} = \dfrac{y-4}{5} = \dfrac{z-5}{6}$ is $\vec{v} = 4\hat{i} + 5\hat{j} + 6\hat{k}$.

A line is parallel to a plane if its direction vector $\vec{v}$ is perpendicular to the normal vector $\vec{n}$ of the plane, which means the dot product $\vec{v} \cdot \vec{n} = 0$.

Checking the options:

For (7XqPP): $\vec{n} = 3\hat{i} + 4\hat{j} + 5\hat{k}$. $\vec{v} \cdot \vec{n} = (4)(3) + (5)(4) + (6)(5) = 12 + 20 + 30 = 62 \neq 0$.

For (L48YE): $\vec{n} = 1\hat{i} + 1\hat{j} + 1\hat{k}$. $\vec{v} \cdot \vec{n} = (4)(1) + (5)(1) + (6)(1) = 4 + 5 + 6 = 15 \neq 0$.

For (2Ytuu): $\vec{n} = 1\hat{i} - 2\hat{j} + 1\hat{k}$. $\vec{v} \cdot \vec{n} = (4)(1) + (5)(-2) + (6)(1) = 4 - 10 + 6 = 0$.

Since the dot product is zero, the line is parallel to the plane $x - 2y + z = 0$.

Answer: $x-2 y+z=0$