In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

The direction ratios of normal to the plane, $L_{1}: a_{1} x+b_{1} y+c_{1} z=0$, are $a_{1}, b_{1}, c_{1}$ and $L_{2}: a_{1} x+b_{2} y+c_{2} z=0$ are $a_{2}, b_{2}, c_{2}$.

$L_{1} \| L_{2}$, if $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$L_{1} \perp L_{2}$, if $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$

The angle between $L_{1}$ and $L_{2}$ is given by,

$Q=\cos ^{-1}\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}}\right|$

(a) The equations of the planes are $7 x+5 y+6 z+30=0$ and

$3 x-y-10 z+4=0$

Here, $a_{1}=7, b_{1}=5, c_{1}=6$

$a_{2}=3, b_{2}=-1, c_{2}=-10$

$a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=7 \times 3+5 \times(-1)+6 \times(-10)=-44 \neq 0$

Therefore, the given planes are not perpendicular.

$\frac{a_{1}}{a_{2}}=\frac{7}{3}, \frac{b_{1}}{b_{2}}=\frac{5}{-1}=-5, \frac{c_{1}}{c_{2}}=\frac{6}{-10}=\frac{-3}{5}$

It can be seen that, $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

Therefore, the given planes are not parallel.

The angle between them is given by,

$Q=\cos ^{-1}\left|\frac{7 \times 3+5 \times(-1)+6 \times(-10)}{\sqrt{(7)^{2}+(5)^{2}+(6)^{2}} \times \sqrt{(3)^{2}+(-1)^{2}+(-10)^{2}}}\right|$

$=\cos ^{-1}\left|\frac{21-5-60}{\sqrt{110} \times \sqrt{110}}\right|$

$=\cos ^{-1} \frac{44}{110}$

$=\cos ^{-1} \frac{2}{5}$

(b) The equations of the planes are $2 x+y+3 z-2=0$ and $x-2 y+5=0$

Here, $a_{1}=2, b_{1}=1, c_{1}=3$ and $a_{2}=1, b_{2}=-2, c_{2}=0$

$\therefore a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=2 \times 1+1 \times(-2)+3 \times 0=0$

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are $2 x-2 y+4 z+5=0$ and $3 x-3 y+6 z-1=0$

Here, $a_{1}=2, b_{1}-2, c_{1}=4$ and $a_{2}=3, b_{2}=-3, c_{2}=6 \quad a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=2 \times 3+(-2)(-3)+4 \times 6=6+6+24=36 \neq 0$

Thus, the given planes are not perpendicular to each other.

$\frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{-2}{-3}=\frac{2}{3}$ and $\frac{c_{1}}{c_{2}}=\frac{4}{6}=\frac{2}{3}$

$\therefore \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

Thus, the given planes are parallel to each other.

(d) The equations of the planes are $2 x-y+3 z-1=0$ and $2 x-y+3 z+3=0$

Here, $a_{1}=2, b_{1}=-1, c_{1}=3$ and $a_{2}=2, b_{2}=-1, c_{2}=3$

$\frac{a_{1}}{a_{2}}=\frac{2}{2}=1, \frac{b_{1}}{b_{2}}=\frac{-1}{-1}=1$ and $\frac{c_{1}}{c_{2}}=\frac{3}{3}=1$

$\therefore \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

Thus, the given lines are parallel to each other.

(e) The equations of the given planes are $4 x+8 y+z-8=0$ and $y+z-4=0$

Here, $a_{1}=4, b_{1}=8, c_{1}=1$ and $a_{2}=0, b_{2}=1, c_{2}=1$

$a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=4 \times 0+8 \times 1+1=9 \neq 0$

Therefore, the given lines are not perpendicular to each other.

$\frac{a_{1}}{a_{2}}=\frac{4}{0}, \frac{b_{1}}{b_{2}}=\frac{8}{1}=8, \frac{c_{1}}{c_{2}}=\frac{1}{1}=1$

$\therefore \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq, \frac{c_{1}}{c_{2}}$

Therefore, the given lines are not parallel to each other.

The angle between the planes is given by,

$Q=\cos ^{-1}\left|\frac{4 \times 0+8 \times 1+1 \times 1}{\sqrt{4^{2}+8^{2}+1^{2}} \times \sqrt{0^{2}+1^{2}+1^{2}}}\right|=\cos ^{-1}\left|\frac{9}{9 \times \sqrt{2}}\right|=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}$