(a) Direction ratios of the normal to the plane are $0,0,1$
$$
\begin{aligned}
&\Rightarrow \mathrm{a}=0, \mathrm{~b}=0, \mathrm{c}=1 \\
&\sqrt{a^2+b^2+c^2}=\sqrt{0^2+0^2+1^2}=1
\end{aligned}
$$
$$
\therefore \quad \frac{0 x}{1}+\frac{0 y}{1}+\frac{z}{1}=\frac{2}{2}
$$
Comparing with $l \mathrm{x}+\mathrm{my}+\mathrm{nz}=\mathrm{p} ; p=\frac{2}{1}=2$
(b) $x+y+z=1$
Direction ratios of the normal to the plane are:
$\mathrm{a}=1, \mathrm{~b}=1, \mathrm{c}=1$ Now $\sqrt{a^2+b^2+c^2}=\sqrt{1^2+1^2+1^2}=\sqrt{3}$
$\therefore \quad \frac{x}{\sqrt{3}}+\frac{y}{\sqrt{3}}+\frac{z}{\sqrt{3}}=\frac{1}{\sqrt{3}}$
Comparing with $l \mathrm{x}+\mathrm{my}+\mathrm{nz}=\mathrm{p} ; \mathrm{p}=\frac{1}{\sqrt{3}}$
(c) $2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}=5 ;$ Now $\mathrm{a}=2 ; \mathrm{b}=3 ; \mathrm{c}=-1$
$\therefore \quad \sqrt{a^2+b^2+c^2}=\sqrt{2^2+3^2+(-1)^2}=\sqrt{14}$
$\therefore \quad \frac{2 x}{\sqrt{14}}+\frac{3 y}{\sqrt{14}}-\frac{z}{\sqrt{14}}=\frac{5}{\sqrt{14}}$
Comparing with $l \mathrm{x}+\mathrm{my}+\mathrm{nz}=\mathrm{p} ; \mathrm{p}=\frac{5}{\sqrt{14}}$
(d) $5 y+8=0 \Rightarrow 5 y=-8 \Rightarrow-5 y=8$
$\begin{aligned} & \text { Now } \mathrm{a}=0 ; \mathrm{b}=-5 ; \mathrm{c}=0 \\ \therefore \quad \sqrt{a^2+b^2+c^2}=\sqrt{0^2+(-5)^2+0^2}=5 \end{aligned}$
$\therefore \quad \frac{0 x}{5}+\frac{-5 y}{5}+\frac{0 z}{5}=\frac{8}{5}$
Comparing with $l \mathrm{x}+\mathrm{my}+\mathrm{nz}=\mathrm{p} ; \therefore \mathrm{p}=\frac{8}{5}$