How do I find the image of a point in a line in 3D space? The question is, 

find the image of the point $(1, 6, 3)$ in the line $$\frac x1 = \frac {y-1}{2} = \frac {z-2}{3}$$

I want to know the general equation to find the image of a point in a line.
EDIT
By image, I mean a point which is diametrically opposite to the point, on the opposite side of the line

Let $Q(a,b,c)$ be the image of $P(1,6,3)$ in the given line.Then direction ratios of $PQ$ are $(a-1,b-6,c-3)$
As the given line and $PQ$ will be perpendicular to each other.So $a_1a_2+b_1b_2+c_1c_2=0$ is the condition of perpendicularity
$1(a-1)+2(b-6)+3(c-3)=0$
$a+2b+3c=22..........(1)$
Now the mid point of $PQ(\frac{a+1}{2},\frac{b+6}{2},\frac{c+3}{2})$ will lie on the line.
Write the equation of line in parametric form
$x=t,y=2t+1,z=3t+2$
So $\frac{a+1}{2}=t,\frac{b+6}{2}=2t+1,\frac{c+3}{2}=3t+2$
$a=2t-1,b=4t-4,c=6t+1$
Put this in equation $(1)$,find $t$ and hence find $(a,b,c)$
