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

(1, 0, 7)

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 given by $\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-2}{3}= λ$

i.e. $ x = λ, y = 2λ + 1, z = 3λ + 2$

So, let the co-ordinates of L be

$(λ , 2λ + 1, 3λ + 2)$

$∴ \vec{PL} = ( λ - 1) \hat{i} + (2 λ - 5) \hat{j} + (3 λ - 1) \hat{k}.$

The given line is parallel to the vector $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}.$

$⇒\vec{PL}.\vec{b} = 0 $

$⇒ (λ -1)  ×1 + ( 2λ-5) ×2 + ( 3λ - 1) × 3 = 0 ⇒ λ= 1$

So, coordinates of L are (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 mid-point of PQ.

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

$⇒x_1 = 1, y_1 = 0 $ and $z_1 = 7 $

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