Find the coordinates of the image of the point $(1, 6, 3)$ with respect to the line $\vec{r} = (\hat{j} + 2\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$; where '$\lambda$' is a scalar. Also, find the distance of the image from the Y-axis.

Image: $(1, 0, 7)$; Distance: $\sqrt{50}$ units

The correct answer is Option (4) → Image: $(1, 0, 7)$; Distance: $\sqrt{50}$ units ##

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

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

$\lambda$ is a scalar, i.e., $x = \lambda, y = 2\lambda + 1$ and $z = 3\lambda + 2$.

Let the coordinates of $L$ be $(\lambda, 2\lambda + 1, 3\lambda + 2)$.

So, direction ratios of $PL$ are $\lambda - 1, 2\lambda + 1 - 6$ and $3\lambda + 2 - 3$, i.e., $\lambda - 1, 2\lambda - 5$ and $3\lambda - 1$.

Direction ratios of the given line are $1, 2$ and $3$, which is perpendicular to $PL$.

Therefore, $(\lambda - 1)1 + (2\lambda - 5)2 + (3\lambda - 1)3 = 0$

$\Rightarrow 14\lambda - 14 = 0 \Rightarrow \lambda = 1$

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 $PQ$.

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

$\Rightarrow x_1 = 1, y_1 = 0$ and $z_1 = 7$.

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

Now, the distance of the point $(1, 0, 7)$ from the Y-axis is $\sqrt{1^2 + 7^2} = \sqrt{50}$ units.