In each of the following cases,determine the direction cosines of the normal to the plane and the distance from the origin: $2x + 3y - z = 5$.

(A) The given equation of the plane is $2x + 3y - z = 5$ .............$(1)$
The direction ratios of the normal to the plane are $a = 2, b = 3, c = -1$.
The magnitude of the normal vector is $\sqrt{a^2 + b^2 + c^2} = \sqrt{(2)^2 + (3)^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.
To convert the equation into the normal form $lx + my + nz = d$,we divide both sides of equation $(1)$ by $\sqrt{14}$:
$\frac{2}{\sqrt{14}}x + \frac{3}{\sqrt{14}}y - \frac{1}{\sqrt{14}}z = \frac{5}{\sqrt{14}}$.
Comparing this with the standard normal form $lx + my + nz = d$,where $(l, m, n)$ are the direction cosines of the normal and $d$ is the distance from the origin:
The direction cosines are $\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}$.
The distance from the origin is $\frac{5}{\sqrt{14}}$ units.