Let P(1, - 2, 3).
The d,r.'s of \(\frac{x - 1}2 = \frac{y - 3}3 = \frac {z + 2}{-6}\) are 2, 3, -6.
Since parallel lines have proportionate d.r.'s so, equation of the line through P(1, -2, 3) and parallel to the given line is:
\(\frac{x - 1}2 = \frac{y +2}3 = \frac {z -3}{-6} = \lambda\) .....(i)
Co-ordinates of any random point on line (i) is Q(2λ + 1, 3λ - 2, 3 - 6λ).
If Q lies on the given equation of plane x - y + z = 5 then, we have
(2λ + 1) - (3λ - 2) + (3 - 6λ) = 5
or λ = \(\frac 17\)
So, co-ordinates of the point Q are Q\(\left(\frac 97, - \frac{11}7, \frac {15}7\right)\).
\(\therefore\) Required distance
PQ = \(\sqrt{\left(\frac 97 - 1\right)^2+\left(-\frac{11}7 + 2\right)^2 + \left(\frac{15}7 - 3\right)^2}\)
= 1 unit