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Q.
The image of the point (3, -1, 11) in the line $\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4} $ is
Three Dimensional Geometry
Solution:
Any point on the given line is (2t, 2 + 3t, 3 + 4t) .
it is the foot of perpendicular from (3, -1, 11)
if $(2t - 3) \times 2 + (2 + 3t +1) \times 3 + (3 + 4t -11) \times 4 = 0$
$\Rightarrow \, t = 1 $
So, the coordinates of foot of perpendicular are (2, 5, 7)
Now, the point (a, b, c) will be the image of the point (3, -1, 11) in the given line if
$\frac{3+a}{2} = 2, \frac{b-1}{2} = 5; $
$ \frac{c+11}{2} = 7 \Rightarrow a = 1, b = 11 , c = 3 $