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

(D) Let $P(1, 6, 3)$ be the given point and let $L$ be the foot of the perpendicular from $P$ to the given line.
The coordinates of a general point on the given line are $\frac{x-0}{1} = \frac{y-1}{2} = \frac{z-2}{3} = \lambda$,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 $(\lambda - 1, 2\lambda + 1 - 6, 3\lambda + 2 - 3)$,which simplifies to $(\lambda - 1, 2\lambda - 5, 3\lambda - 1)$.
The direction ratios of the given line are $(1, 2, 3)$. Since $PL$ is perpendicular to the line,the dot product of their direction ratios must be zero:
$1(\lambda - 1) + 2(2\lambda - 5) + 3(3\lambda - 1) = 0$
$\lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0$
$14\lambda - 14 = 0 \Rightarrow \lambda = 1$.
Substituting $\lambda = 1$ into the coordinates of $L$,we get $L(1, 2(1) + 1, 3(1) + 2) = (1, 3, 5)$.
Let $Q(x_1, y_1, z_1)$ be the image of $P(1, 6, 3)$ in the given line. Then $L$ is the midpoint of $PQ$. Therefore:
$\frac{x_1 + 1}{2} = 1 \Rightarrow x_1 = 1$
$\frac{y_1 + 6}{2} = 3 \Rightarrow y_1 = 0$
$\frac{z_1 + 3}{2} = 5 \Rightarrow z_1 = 7$
Hence,the image of $(1, 6, 3)$ in the given line is $(1, 0, 7)$.