Image of point – 3 D Geometry – JEE Main 24 January 2025 Shift 2

Solution:

Since the point $R(5, p, q)$ lies on the line $L$, its coordinates must satisfy the line equation.
$$ \frac{5-1}{2} = \frac{p+1}{3} = \frac{q}{4} $$
$$ 2 = \frac{p+1}{3} \Rightarrow p = 5 $$
$$ 2 = \frac{q}{4} \Rightarrow q = 8 $$
So, $R$ is $(5, 5, 8)$.

Next, we find the foot of the perpendicular, $T$, from $Q(7, -2, 5)$ to the line $L$. Let $T = (2\lambda+1, 3\lambda-1, 4\lambda)$.
The vector $\vec{QT} = (2\lambda-6)\hat{i} + (3\lambda+1)\hat{j} + (4\lambda-5)\hat{k}$ is perpendicular to the line’s direction vector $2\hat{i} + 3\hat{j} + 4\hat{k}$.
$$ 2(2\lambda-6) + 3(3\lambda+1) + 4(4\lambda-5) = 0 $$
$$ 4\lambda – 12 + 9\lambda + 3 + 16\lambda – 20 = 0 $$
$$ 29\lambda = 29 \Rightarrow \lambda = 1 $$
The coordinates of $T$ are $(3, 2, 4)$.

Since $P$ is the image of $Q$ in the line, the triangle $PQR$ is isosceles with $PQ$ as the base and $RT$ as the altitude. The area is given by:
$$ \text{Area} = \frac{1}{2} \times \text{Base}(PQ) \times \text{Height}(RT) = \frac{1}{2} \times (2QT) \times RT = QT \times RT $$
Calculating the squared distances:
$$ QT^2 = (3-7)^2 + (2+2)^2 + (4-5)^2 = 16 + 16 + 1 = 33 $$
$$ RT^2 = (5-3)^2 + (5-2)^2 + (8-4)^2 = 4 + 9 + 16 = 29 $$

The square of the area is:
$$ \text{Area}^2 = QT^2 \times RT^2 = 33 \times 29 = 957 $$

Ans. (957)