In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angles between them.
$2x + y + 3z - 2 = 0$ and $x - 2y + 5 = 0$

(B) The direction ratios of the normal to the plane $L_1: a_1x + b_1y + c_1z + d_1 = 0$ are $(a_1, b_1, c_1)$ and for $L_2: a_2x + b_2y + c_2z + d_2 = 0$ are $(a_2, b_2, c_2)$.
$L_1 \parallel L_2$ if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
$L_1 \perp L_2$ if $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
The angle $\theta$ between $L_1$ and $L_2$ is given by $\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
The equations of the planes are $2x + y + 3z - 2 = 0$ and $x - 2y + 0z + 5 = 0$.
Here,$a_1 = 2, b_1 = 1, c_1 = 3$ and $a_2 = 1, b_2 = -2, c_2 = 0$.
Calculating the dot product of the normal vectors: $a_1a_2 + b_1b_2 + c_1c_2 = (2)(1) + (1)(-2) + (3)(0) = 2 - 2 + 0 = 0$.
Since the dot product is $0$,the given planes are perpendicular to each other.