Let the system of linear equations
 $\begin{array}{l}-\mathrm{x}+2\mathrm{y}-9\mathrm{z}=7\\ -\mathrm{x}+3\mathrm{y}+7\mathrm{z}=9\\ -2\mathrm{x}+\mathrm{y}+5\mathrm{z}=8\\ -3\mathrm{x}+\mathrm{y}+13\mathrm{z}=\mathrm{\lambda }\end{array}$
has a unique solution $\mathrm{x}=\mathrm{\alpha }, \mathrm{y}=\mathrm{\beta }, \mathrm{z}=\mathrm{\gamma }$ then distance of the point $(\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma })$ from the plane 2x - 2y + z = $\mathrm{\lambda }$ is

From first three equations $\mathrm{\Delta }=-71, {\mathrm{\Delta }}_1=213, {\mathrm{\Delta }}_2=-142, {\mathrm{\Delta }}_3=0$
 from Cramer 's rule $\mathrm{x}=\mathrm{\alpha }=-3, \mathrm{y}=\mathrm{\beta }=2, \mathrm{z}=\mathrm{\gamma }=0,$
Substitute in (iv)  3 × 3 + 2 = $\lambda$
$\lambda$ = 11
 Distance from (-3, 2, 0) to the plane 2x - 2y + z = 11 = 0 
$d=\left|\frac{-6-4-11}{\sqrt{2^2+2^2+1}}\right|=\frac{21}{3}=7$