Let the equation of the plane passing through the line x−2y−z−5=0=x+y+3z−5 and parallel to the line x+y+2z−7=0=2x+3y+z−2 be ax+by+cz=65. Then the distance of the point (a,b,c) from the plane 2x+2y−z+16=0 is:
Answer: 9
Solutions · 1
Official solution
Direction of the second given line: (1,1,2)×(2,3,1)=(−5,3,1).
Family through the first line: (1+k)x+(k−2)y+(3k−1)z−(5+5k)=0. Its normal must be perpendicular to (−5,3,1):
\[-5(1+k)+3(k-2)+(3k-1)=0 \implies k-12=0 \implies k=12.\]
\[\text{Plane: } 13x+10y+35z=65 \implies (a,b,c)=(13,10,35).\]
\[d=\frac{|2(13)+2(10)-35+16|}{\sqrt{4+4+1}}=\frac{27}{3}=9.\]