Three Dimensional Geometry Part 2

Given equation of line

x−2

α

=

y−2

−5

=

z+2

2

...(i)
and plane x+3y−2z+β=0 ...(ii)
Line (i) pases through (2,2,−2)
which lies on plane (ii).
∴2+6+4+β=0
⇒ β = −12
Also, given line is perpendicular to normal of the plane
α(1)−5(3)+2(−2)=0
⇒ α = 19
∴α+β=7