The direction cosines of the normal to the plane $x-y+z=4$ are :

$\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$

for plane : Ax + By + Cz = D  →  its normal vector is  $A\hat{i} + B\hat{j} + C\hat{k}$

so for x - y + z = 4 ⇒  $\vec{n} = \hat{i} - \hat{j} + \hat{k}$

as $|\vec{n}|$

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

= $\sqrt{3}$

$\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$

So direction cosins are $\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$