Equation of the line through (1,−2,3) parallel to the line ‌
x
2
=‌
y
3
=‌
z−1
−6
is ‌
x−1
2
=‌
y+2
3
=‌
z−3
−6
=r (say) . . . (i) Then, any point on Eq. (i) is (2r+1,3r−2,−6r+3) If this point lies on the plane x−y+z=5, then (2r+1)−(3r−2)+(−6r+3)=5 ⇒−7r+6=5⇒r=‌