A straight line and a plane will be perpendicular when the governing vector of the straight line is parallel to the normal vector of the plane. Since we know that two vectors are parallel if and only if they are linearly dependent, we will only have to verify if its components are proportional.
A straight line $$r$$ with governing vector $$\vec{v}=(v_1,v_2,v_3)$$ and a plane $$\pi$$ with normal vector $$\vec{n}=(A,B,C)$$ are perpendicular if and only if $$\vec{v}$$ and $$\vec{n}$$ are linearly dependent:
$$$r \ \text{ perpendicular to } \ \pi \Leftrightarrow \ \dfrac{v_1}{A}=\dfrac{v_2}{B}=\dfrac{v_3}{C}$$$
The straight line $$r: (x, y, z) = (2, 0, -3) + k\cdot(-1, 3, 5)$$ and the plane $$\pi: x - 3y - 5z + 2 = 0$$ are perpendicular since the governing vector $$\vec{v} = (-1, 3, 5)$$ and the normal vector $$\vec{n} = (1, -3, -5)$$ verify: $$$ \dfrac{-1}{1}=\dfrac{3}{-3}=\dfrac{5}{-5}$$$