Let the system of linear equations

$-x+2 y-9 z=7$

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

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

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

has a unique solution $x=\alpha, y=\beta, z=\gamma$. Then the distance of the point

$(\alpha, \beta, \gamma)$ from the plane $2 x-2 y+z=\lambda$ is :

$$ \begin{aligned} & -x+2 y-9 z=7-(1) \\\\ & -x+3 y-7 z=9-(2) \\\\ & -2 x+y+5 z=8-(3) \\\\ & (2)-(1) \\\\ & y+16 z=2 \quad(4) \\\\ & (3)-2 \times(1) \\\\ & -3 y+23 z=-6-(5) \\\\ & 3 \times(4)+(5) \\\\ & 71 z=0 \Rightarrow z=0 \\\\ & \quad y=2 \\\\ & (-3,2,0) \rightarrow(\alpha, \beta, \gamma) \\\\ & \text { Put in }-3 x+y+13 z=\lambda \\\\ & \lambda=9+2=11 \end{aligned} $$ <br/><br/>The equation of the plane is: <br/><br/>$2x - 2y + z = \lambda$ <br/><br/>We already know that $\lambda = 11$. So the equation of the plane becomes: <br/><br/>$2x - 2y + z = 11$ <br/><br/>Now, let's find the distance $d$ of the point $(\alpha, \beta, \gamma) = (-3, 2, 0)$ from the plane using the formula : <br/><br/>$d = \frac{|Ax + By + Cz + D|}{\sqrt{A^2 + B^2 + C^2}}$ <br/><br/>Plugging in the values for the point and the plane equation, we get: <br/><br/>$$ d = \frac{|2(-3) - 2(2) + 0 - 11|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|-6 - 4 - 11|}{\sqrt{4 + 4 + 1}} = \frac{21}{\sqrt{9}} = 7 $$ <br/><br/>So, the distance of the point $(-3, 2, 0)$ from the plane $2x - 2y + z = 11$ is 7.