The place $x-2 y+z=0$ is parallel to the line

A line with direction vector $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$ is parallel to a plane with normal vector $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$ if and only if $\vec{n} \cdot \vec{b} = 0$.

The normal vector to the plane $x - 2y + z = 0$ is $\vec{n} = 1\hat{i} - 2\hat{j} + 1\hat{k}$.

We check the dot product for each option:

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

Since the dot product is zero, the line is parallel to the plane. We also check if the line lies in the plane by testing a point on the line, such as $(3, 4, 3)$. Substituting into the plane equation: $3 - 2(4) + 3 = 3 - 8 + 3 = -2 \neq 0$. Thus, the line is parallel to the plane and does not lie in it.

Answer: $\dfrac{x-3}{4}=\dfrac{y-4}{5}=\dfrac{z-3}{6}$