Given a.3+(−4a)(−1)+(−7)2b=0 and ab−4a2+14=0...... (2) ⇒a2=4 and b2=1 ∴L≡‌
x+1
5
=‌
y−2
3
=‌
z
1
=λ (say) ⇒ General point on line is (5λ−1,3λ+2,λ) for finding point of intersection with x−y+z=0 we get (5λ−1)−(3λ+2)+(λ)=0 ⇒3λ−3=0⇒λ=1 ∴ Point at intersection (4,5,1) ∴α+β+γ=4+5+1=10