Let $P(1,6,3)$ be the given point and $L$ be the foot of perpendicular from $P$ to the given line. The coordinates of a general point on the given line are

$\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda$   (say)

i.e. $x=\lambda$

$y=2\lambda +1, z=3\lambda +2$ 

If the coordinates of $L$ are $(\lambda ,2\lambda +1,3\lambda +2)$, then the direction ratios of $PL$ are $(\lambda -1,2\lambda -5,3\lambda -1)$.

Since, the direction ratios of given line which is perpendicular to $PL$, are $1,2$ and $3$. Therefore, $(\lambda -1)1+(2\lambda -5)2+(3\lambda -1)3=0$,which gives $\lambda =1$.

Hence, coordinates of $L$ are $(1,3,5)$.

Let $Q\left(x_1,y_1,z_1\right)$ be the image of $P(1,6,3)$ in the given line. Then, $L$ is the mid-point of $PQ$.

Therefore, $\frac{x_1+1}{2}=1,\frac{y_1+6}{2}=3,\frac{z_1+3}{2}=5$

$\Rightarrow x_1=1,y=0,z=7$

Hence, the image of $(1,6,3)$ in the given line is $(1,0,7)$.