Let the system of linear equations

$–x+2y-9z=7$

$-x+3y+7z=9$

$-2x+y+5z=8$

$-3x+y+13z=\lambda$

has a unique solution $x=\alpha , y=\beta , z=\gamma$. Then the distance of the point $\left(\alpha , \beta , \gamma \right)$ from the plane $2x-2y+z=\lambda$ is

Given,

System of linear equation,

$–x+2y-9z=7 .......\left(\mathrm{i}\right)$

$–x+3y+7z=9 .......\left(\mathrm{ii}\right)$

$–2x+y+5z=8 ......\left(\mathrm{iii}\right)$

$–3x+y+13z=\lambda ......\left(\mathrm{iv}\right)$

Now solving, equations $\left(\mathrm{i}\right), \left(\mathrm{ii}\right) \& \left(\mathrm{iii}\right)$ we get,

$x=–3, y=2, z=0$

Now substituting in equation $\left(\mathrm{iv}\right)$ we get,

$3\times 3+2=\lambda$

$\Rightarrow \lambda =11$

Now finding the distance of Point $\left(–3, 2, 0\right)$ from the plane $2x-2y+z=11$ we get,

$d=\left|\frac{-6-4-11}{\sqrt{2^2+2^2+1}}\right|=\frac{21}{3}=7$