Find the image of the line `(x-1)/(9)=(y-2)/(-1)=(z+3)/(-3)` in the plane `3x-3y+10z-26=0.`

Explanation

`" "(x-1)/(9)=(y-2)/(-1)=(z+3)/(-3)" "`(i)
`" " 3x-3y+10z-26=0" "`(ii)
The direction ratios of the line are 9, -1 and -3 and direction ratios of the normal to the given plane are 3, -3 and 10.
Since `9*3+(-1)(-3)+(-3)10=0` and the point (1, 2, -3) of line (i) does not lie in plane (ii) for `3*1-3*2+10*(-3)-26 ne 0,` line (i) is parallel to plane (ii). Let `A'` be the image of point `A(1, 2, -3)` in point (ii). Then the image of the line (i) in the plane (ii) is the line through `A'` and parallel to the line (i).
Let point `A'` be (`p, q, r`). Then
`" "(p-1)/(3)=(q-2)/(-3)=(r+3)/(10)=-((3(1)-3(2)+10(-3)-26)/(9+9+100)=(1)/(2)`
The point is `A'(5//2, 1//2, 2)`.
The equation of line `BA'` `(x-(5//2))/(9)=(y-(1//2))/(-1)=(z-2)/(-3)`