Let αx+βy+γz=1 be the equation of a plane passing through the point (3,−2,5) and perpendicular to the line joining the points (1,2,3) and (−2,3,5). Then the value of αβγ is equal to:
Answer: 6
Solutions · 1
Official solution
Direction of the line: (−3,1,2). Plane: −3x+y+2z=d, through (3,−2,5): −9−2+10=−1=d.
\[-3x+y+2z=-1 \implies 3x-y-2z=1.\]
So α=3,β=−1,γ=−2.
\[\alpha\beta\gamma=3\times(-1)\times(-2)=6.\]