The direction ratios of the line, given by the planes `x - y + z - 5 = 0, x - 3y - 6 = 0 `are

Explanation

Any point on the line `(x-1)/(1)= (y+1)/(-2)= (z-2)/(3) = t` is `(t+1, -2t-1, 3t+2)`, which lies on the given plane if `t+1+ 2t+1+6t+4-3=0 or t= -1//3`.
Hence, the point of intersection of the line and the plane is `P(2//3, -1//3, 1)`.
Also, if the foot of the perpendicular from `A(1, -1, 2)` on the plane is `Q(x, y, z)`, then
`" "(x-1)/(1)=(y+1)/(-1)= (z-2)/(2)`
`" "=-((1+1+4-3))/(1+1+4) = -(1)/(2)`
Therefore, `Q(x, y, z)` is `Q(1//2, -1//2, 1)`.
Hence, the direction ratios of `PQ` are `(2)/(3)-(1)/(2), - (1)/(3)+ (1)/(2) and 1-1 or (1)/(6), (1)/(6) and 0`.
If the image of point `A(1, -1, 2)` in the plane is `R`, then `Q` is the midpoint of `AR`. Therefore, point `R` is `(0, 0, 0)`.
Hence, the direction ratios of `PR` or the image of the line in the plane are `2//3, -1//3 and 1`.