If the equation of the plane containing the line x+2y+3z−4=0=2x+y−z+5 and perpendicular to the plane r=(i^−j^)+λ(i^+j^+k^)+μ(i^−2j^+3k^) is ax+by+cz=4, then (a−b+c) is equal to:
Solutions · 1
Official solution
Direction of line: n1×n2 where n1=(1,2,3),n2=(2,1,−1): (−5,7,−3).
Normal of the given plane: (1,1,1)×(1,−2,3)=(5,−2,−3).
\[\vec n_1\times\vec n_2\] combined with the perpendicularity condition and a point on the intersection line (0,−511,514) gives the required plane:
27x+30y+25z=4.
\[a=27,\ b=30,\ c=25 \implies a-b+c=22.\]