If the equation of the plane that contains the point (−2,3,5) and is perpendicular to each of the planes 2x+4y+5z=8 and 3x−2y+3z=5 is αx+βy+γz+97=0, then α+β+γ=:
Solutions · 1
Official solution
Normal ∝(2,4,5)×(3,−2,3)=(22,9,−16).
\[22(x+2)+9(y-3)-16(z-5)=0 \implies 22x+9y-16z+97=0.\]
\[\alpha=22,\ \beta=9,\ \gamma=-16 \implies \alpha+\beta+\gamma=15.\]