In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angles between them.
$2x - y + 3z - 1 = 0$ and $2x - y + 3z + 3 = 0$

(A) 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 $\theta = \cos^{-1} \left( \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} \right)$.
The equations of the planes are $2x - y + 3z - 1 = 0$ and $2x - y + 3z + 3 = 0$.
Here,$a_1 = 2, b_1 = -1, c_1 = 3$ and $a_2 = 2, b_2 = -1, c_2 = 3$.
Calculating the ratios: $\frac{a_1}{a_2} = \frac{2}{2} = 1$,$\frac{b_1}{b_2} = \frac{-1}{-1} = 1$,and $\frac{c_1}{c_2} = \frac{3}{3} = 1$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = 1$,the given planes are parallel to each other.