Given,
System of linear equation,
–x+2y−9z=7.......(i)
–x+3y+7z=9.......(ii)
–2x+y+5z=8......(iii)
–3x+y+13z=λ......(iv)
Now solving, equations (i),(\mathrm{ii})&(\mathrm{iii}) we get,
x=–3,y=2,z=0
Now substituting in equation (iv) we get,
3×3+2=λ
⇒λ=11
Now finding the distance of Point (–3,2,0) from the plane 2x−2y+z=11 we get,
d=∣22+22+1−6−4−11∣=321=7