The equation of any line through (1, 0, –3) and parallel to the line (x – 2)/2 = (y + 2)/3 = (6 – z)6 is
Any point on the line (i) is (2λ + 1, 3λ, – 6 – 3λ)
which is lies in the plane x – y – z = 9. So,
(2λ + 1) – (3λ) – (–6 – 3λ) = 9
2λ + 7 = 9
2λ = 2
λ = 1
Thus, the point is (3, 3, –9).
Hence, the required distance