Given line is

It can be written in cartesian from as

Let Q(\(\alpha, \beta, \gamma\)) be the foot of perpendicular drawn from P(5, 4, 2) to the line (i) and P' (\(x_1, y_1,z_1\)) be the image of P on the line(i)


Parallel vector of line (i) \(\vec{b}=2\hat{i}+3\hat{j}-\hat{k}.\)


[Putting value of \(\alpha, \beta, \gamma\) from (ii)]

Hence the coordinates of foot of perpendicular Q are \((2\times 1-1,\,3\times 1+3,\, -1+1),\) i.e., (1, 6, 0)
\(\therefore\) Length of perpendicular


Also, since Q is mid-point of PP'

Therefore required image is (-3, 8, -2).