The plane passing through the point ($-2, -2, 2$) and containing the line joining the points 

($1, 1, 1$) and ($1, -1, 2$) makes intercepts on the co-ordinate axes, the  absolute value of the sum of intercepts  is 

Equation of any plane passing through $(-2,-2, 2)$ is a $(x +2) + b (y+2) +c(z-2) =0$

since it contains the line joining the points ($1, 1, 1$) and($1,-1, 2$), it contains these points

as well so that $3a+3b-c=0$ and $3a+b+0=0$

on solving we get $\frac{\mathrm{a}}{1}=\frac{\mathrm{b}}{-3}=\frac{\mathrm{c}}{-6}$ and thus

the equation of the plane is 

$\begin{array}{r}x+2-3(y+2)-6(z-2)=0\\ \end{array}$

or $x-3y-6z+8=0$

or $\frac{x}{-8}+\frac{y}{8/3}+\frac{z}{8/6}=1$

The requd. sum $=-8+\frac{8}{3}+\frac{8}{6}=-4$