Step 1:

The given equation is $2x+3y-z=5$---(1)

The direction ratios of the normal are $(2,3,-1)$

Therefore $\sqrt{2^2+3^2+(-1)^2}=\sqrt{14}$

Step 2:

Divide equ(1) on both sides by $\sqrt{14}$

$\large\frac{2}{\sqrt{14}}$$x+\large\frac{3}{\sqrt{14}}$$y-\large\frac{z}{\sqrt{14}}=\frac{5}{\sqrt{14}}$

This equation of the form $lx+my+nz=d$,where $l,m$ and $n$ are the direction cosines of normal to the plane and $d$ is the distance of normal from the origin.

Therefore the direction cosines are $\big(\large\frac{2}{\sqrt {14}},\large\frac{3}{\sqrt {14}},\large\frac{-1}{\sqrt {14}}\big)$ and distance from the origin is $\large\frac{5}{\sqrt {14}}$units.