Equation of plane containing the line 4ax−y+5z−7a=0=2x−5y−z−3 can be written as 4ax−y+5z−7a+λ(2x−5y−z−3)=0 (4a+2λ)x−(1+5λ)y+(5−λ)z−(7z+3λ)=0 Which is coplanar with the line
x−4
1
=
y+1
−2
=
z
1
4(4a+2λ)+(1+5λ)−(7a+3λ)=0 9a+10λ+1=0.....(1) (4a+2λ)1+(1+5λ)2+5−λ=0 4a+11λ+7=0.....(2) a=1,λ=−1 Equation of plane is x+2y+3z−2=0 Intersection with the line
x−3
7
=
y−2
−1
=
z−3
−4
(7t+3)+2(−t+2)+3(−4t+3)−2=0 −7t+14=0 t=2 So, the required point is (17,0,−5) α+β+γ=12