Let P be the plane, passing through the point (1,−1,−5) and perpendicular to the line joining the points (4,1,−3) and (2,4,3). Then the distance of P from the point (3,−2,2) is :
Solutions · 1
Official solution
Normal direction: (2−4,4−1,3+3)=(−2,3,6).
\[P:\ 2(x-1)-3(y+1)-6(z+5)=0 \implies 2x-3y-6z=35.\]
\[d=\frac{|2(3)-3(-2)-6(2)-35|}{\sqrt{4+9+36}}=\frac{35}{7}=5.\]