Find the image of point `(1,2,3)` in the line `(x-6)/3=(y-7)/2=(z-7)/(-2)dot`

Explanation

Let the image of point `A(1, 2, 3)` in the line `l` whose equation is
`" " (x-6)/(3)=(y-7)/(2)=(z-7)/(-2)=k` (say) `" "` (i)
be `A'`. Then `AA'` is perpendicular to `l` and the point of intersection of `AA'` and `l` is the midpoint of `AA'`. Note that `M` is the foot of perpendicular from `A` to `l`.
The coordinates of any point on the given line are of the form `(3k +6, 2k+7, -2k+7)`. Therefore, the coordinates of `M` are `3k+6, 2k+7 and -2k+7` for some value of `k`. The direction ratios of `AM` are `3k+6-1, 2k+7-2 and -2k+7-3 or 3k+5, 2k+5, -2k+4`.
Also, the direction ratios of `l` are 3, 2 and -2.
since `AM bot l, a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0`.
`rArr" "3(3k+5)+2(2k+5)-2(-2k+4)=0`
or `" "17k+17=0 or k=-1`
Thus, the coordinates of `M` are 3, 5 and 9.
Suppose coordinates of `A'` are `x, y andz`,
The coordinates of the midpoint of `AA'` are
`(x+1)/(2), (y+2)/(2) and (z+3)/(2)` .
But the midpoint of `AA'` is (5, 3, 9). Therefore,
`" "(x+1)/(2)=3, (y+2)/(2) =5 and (z+3)/(2)=9`
`rArr" "x=5, y=8, z=15`
Thus, the image of `A` in `l` is (5, 8, 15).