Solution:

(a) $z=2$
It is given that
The eq. of the plane, $z=2$ or $0 x+0 y+z=2 \ldots (1) .$
The direction ratio of the normal $(0,0,1)$
Using the formula,
$\begin{array}{l}
\sqrt\left[(0)^{2}+(0)^{2}+(1)^{2}\right]=\sqrt{1} \\
=1
\end{array}$
Now,
On dividing both sides of the eq. (1) by 1, we obtain
$0 x /(1)+0 y /(1)+z / 1=2$
So this is of the form $I x+m y+n z=d$
Where, I, $\mathrm{m}, \mathrm{n}$ are the direction cosines and $\mathrm{d}$ is the distance
$\therefore$ The direction cosines are $0,0,1$
The distance (d) from the origin is 2 units

(b) $x+y+z=1$
It is given that
The eq. of the plane, $x+y+z=1$…. (1)
The direction ratio of the normal $(1,1,1)$
Using the formula,
$\sqrt\left[(1)^{2}+(1)^{2}+(1)^{2}\right]=\sqrt{3}$
Now,
Dividing both sides of the eq. (1) by $\sqrt{3}$, we obtain
$\mathrm{x} /(\sqrt{3})+\mathrm{y} /(\sqrt{3})+\mathrm{z} /(\sqrt{3})=1 / \sqrt{3}$
So this is of the form $I x+m y+n z=d$
Where, I, $\mathrm{m}, \mathrm{n}$ are the direction cosines and $\mathrm{d}$ is the distance
Therefore the direction cosines are $1 / \sqrt{3}, 1 / \sqrt{3}, 1 / \sqrt{3}$
The distance (d) from the origin is $1 / \sqrt{3}$ units