What are the direction cosines of the equation of the plane 2x + 3y – z = 5?

The equation of the plane, 2x + 3y – z = 5…. (1)

Direction ratio of the normal (2, 3, -1)

By using the formula,
$\sqrt{ }\left[(2{)}^2+(3{)}^2+(-1{)}^2\right]=\sqrt{14}$
Now,
Divide both the sides of equation (1) by  $\sqrt{14}$, we get
$2\mathrm{x}/\left(\sqrt{14}\right)+3\mathrm{y}/\left(\sqrt{14}\right)-\mathrm{z}/\left(\sqrt{14}\right)=5/\sqrt{14}$
So this is of the form lx + my + nz = d
Where, l, m, n are the direction cosines and d is the distance
Therefore, the direction cosines are $2/\sqrt{14},3/\sqrt{14},-1/\sqrt{14}$