The plane passing through the point (–2, –2, 2) and containing the line joining points (1, 1, 1) and (1, –1, 2) makes intercepts of lengths a, b, c respectively on the axes of x, y and z respectively then

Equation of plane passing through (–2, –2, 2) is
l(x+2) + m (y+2) + n(z–2) = 0
Where l, m, n are dr’s of normal to the plane
Since it contains the line joining (1, 1, 1) and (1, –1,2) these points also lie on the planes
$\begin{array}{l}\Rightarrow 3l+3m-n=0\text{ and }3l+m=0\\ \Rightarrow \frac{l}{1}=\frac{m}{-3}=\frac{n}{-6}\end{array}$
$\Rightarrow$ equation of the plane is
(x + 2) – 3(y + 2) – 6(z – 2) = 0 or
$\begin{array}{l}x-3y-6z+8=0\\ \Rightarrow \frac{x}{-8}=\frac{y}{8/3}=\frac{z}{8/6}\\ \Rightarrow a=8,b=8/3,c=8/6\end{array}$