The sum of the intercepts on the coordinate axes of the plane passing through the point $(-2, -2, 2)$ and containing the line joining the points 

(1, -1, 2) and (1, 1, 1), is

Equation of plane is given by 

$\begin{array}{l}\left|\begin{array}{cll}x-(-2)& y-(-2)& z-2\\ -3& -1& 0\\ -3& -3& 1\end{array}\right|=0\\ \Rightarrow \left|\begin{array}{clc}x+2& y+2& z-2\\ -3& -1& 0\\ -3& -3& 1\end{array}\right|=0\\ \Rightarrow (x+2)(-1-0)-(y+2)(-3-0)+(z-2)(9-3)=0\\ \Rightarrow -(x+2)+3(y+2)+6(z-2)=0\\ \Rightarrow -x-2+3y+6+6z-12=0\\ \Rightarrow -x+3y+6z-8=0\Rightarrow x-3y-6z+8=0\\ \Rightarrow \frac{x}{-8}+\frac{y}{8/3}+\frac{z}{8/6}=1\end{array}$

$\therefore$ Sum of intercepts $=-8+\frac{8}{3}+\frac{8}{6}=-4$