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−1
−6
= r (say) ... (1) Then any point on (1) 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) = 2 ⇒ r =