Solution:

(a) $2 x+3 y-z=5$
It is given that
The eq. of the plane, $2 x+3 y-z=5 \ldots$. (1)
The direction ratio of the normal $(2,3,-1)$
Using the formula,
$\left.\sqrt{[}(2)^{2}+(3)^{2}+(-1)^{2}\right]=\sqrt{14}$
Now,
Dividing both sides of eq. (1) by $\sqrt{14}$, we obtain
$2 x /(\sqrt{14})+3 y /(\sqrt{14})-z /(\sqrt{14})=5 / \sqrt{14}$
So this is of the form $I x+m y+n z=d$
Where, I, $m, n$ are the direction cosines and $d$ is the distance
Therefore the direction cosines are $2 / \sqrt{14}, 3 / \sqrt{14},-1 / \sqrt{14}$
Distance (d) from the origin is $5 / \sqrt{14}$ units

(b) $5 y+8=0$
It is given that
The eq. of the plane, $5 y+8=0$
$-5 y=8$ or
$0 x-5 y+0 z=8 \ldots$
The direction ratio of the normal $(0,-5,0)$
Using the formula,
$\begin{array}{l}
\sqrt\left[(0)^{2}+(-5)^{2}+(0)^{2}\right]=\sqrt{25} \\
=5
\end{array}$
Now,
Dividing both sides of eq. (1) by 5 , we obtain
$0 x /(5)-5 y /(5)-0 z /(5)=8 / 5$
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 $0,-1,0$

The distance (d) from the origin is 8/5 units