The coordinates of the image of the point $P(5, 4, 2)$ in the line $\vec r= (-\hat i+3\hat j+ \hat k) +λ(2\hat i +3\hat j-\hat k)$, where is a parameter, is

(-3, 8,-2)

The correct answer is Option (3) → (-3, 8,-2)

Given point: $P(5, 4, 2)$

Line: $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$

Let the foot of perpendicular from $P$ to the line be $Q$

Position vector of point $P$: $\vec{p} = 5\hat{i} + 4\hat{j} + 2\hat{k}$

Point on the line: $\vec{r} = (-1 + 2\lambda)\hat{i} + (3 + 3\lambda)\hat{j} + (1 - \lambda)\hat{k}$

Let this be $\vec{q}$. Vector $\vec{pq} = \vec{q} - \vec{p}$

This vector is perpendicular to the line’s direction vector $\vec{d} = 2\hat{i} + 3\hat{j} - \hat{k}$

So, $(\vec{q} - \vec{p}) \cdot \vec{d} = 0$

Compute:

$[( -1 + 2\lambda - 5)\hat{i} + (3 + 3\lambda - 4)\hat{j} + (1 - \lambda - 2)\hat{k}] \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 0$

$(-6 + 2\lambda)\cdot 2 + (-1 + 3\lambda)\cdot 3 + (-1 - \lambda)\cdot (-1) = 0$

$-12 + 4\lambda -3 + 9\lambda + 1 + \lambda = 0$

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

Substitute $\lambda = 1$ in line equation to get foot of perpendicular $Q$:

$Q = (-1 + 2, 3 + 3, 1 - 1) = (1, 6, 0)$

Foot of perpendicular is $Q(1, 6, 0)$

To get the image of point $P$ in the line, reflect it across $Q$:

Use midpoint formula: $Q = \frac{P + P'}{2} \Rightarrow P' = 2Q - P$

$P' = 2(1, 6, 0) - (5, 4, 2) = (2 - 5, 12 - 4, 0 - 2) = (-3, 8, -2)$