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Let the given line be $L: \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} = \lambda$. Any point on this line can be represented as $P(\lambda, 2\lambda+1, 3\lambda+2)$.
Let $A' (x', y', z')$ be the image of $A(1,6,3)$ in the line $L$. Then the midpoint of $AA'$, say $M$, must lie on the line $L$. The coordinates of $M$ are given by $M(\frac{x'+1}{2}, \frac{y'+6}{2}, \frac{z'+3}{2})$.
Since $M$ lies on the line $L$, we can equate the coordinates of $M$ with the parametric representation of the line: $$ \frac{x'+1}{2} = \lambda, \quad \frac{y'+6}{2} = 2\lambda+1, \quad \frac{z'+3}{2} = 3\lambda+2 $$ This gives us: $$ x' = 2\lambda - 1, \quad y' = 4\lambda - 4, \quad z' = 6\lambda + 1 $$ So, $A' (2\lambda - 1, 4\lambda - 4, 6\lambda + 1)$.
The direction ratios of the line $AA'$ are given by $(2\lambda - 1 - 1, 4\lambda - 4 - 6, 6\lambda + 1 - 3) = (2\lambda - 2, 4\lambda - 10, 6\lambda - 2)$. The direction ratios of the given line $L$ are $(1, 2, 3)$. Since $AA'$ is perpendicular to $L$, the dot product of their direction ratios must be zero: $$ 1(2\lambda - 2) + 2(4\lambda - 10) + 3(6\lambda - 2) = 0 $$ $$ 2\lambda - 2 + 8\lambda - 20 + 18\lambda - 6 = 0 $$ $$ 28\lambda - 28 = 0 $$ $$ \lambda = 1 $$
Substituting $\lambda = 1$ into the coordinates of $A'$, we get: $$ A' (2(1) - 1, 4(1) - 4, 6(1) + 1) = A'(1, 0, 7) $$
The equation of the line joining $A(1,6,3)$ and $A'(1,0,7)$ is given by: $$ \frac{x - 1}{1 - 1} = \frac{y - 6}{0 - 6} = \frac{z - 3}{7 - 3} $$ $$ \frac{x - 1}{0} = \frac{y - 6}{-6} = \frac{z - 3}{4} $$ This implies $x = 1$, and $\frac{y - 6}{-6} = \frac{z - 3}{4}$. Thus, $x = 1$ and $4(y - 6) = -6(z - 3)$, which simplifies to $4y - 24 = -6z + 18$, or $4y + 6z = 42$, or $2y + 3z = 21$.
Final Answer: A'(1, 0, 7) and the equation of the line AA' is x=1, 2y+3z=21
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