Find the direction cosines of the normal to the plane x + 2y + 2z – 4 = 0

Given plane is x + 2y + 2z = 4

compare with ax+by+cz+d=0
Where a = 1, b = 2, c = 2
$\text{ dividing with }\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}=\sqrt{1+4+4}=\sqrt{9}=3 on both sides$
$\frac{1}{3}\mathrm{x}+\frac{2}{3}\mathrm{y}+\frac{2}{3}\mathrm{z}=\frac{4}{3}$
Which is of the form lx + my + nz = P
$\therefore \text{ d.c's of normal to the plane }(\mathrm{l},\mathrm{m},\mathrm{n})=\left(\frac{1}{3},\frac{2}{3},\frac{2}{3}\right)$