3D Geometry – Image of a Point in a Line – JEE Main 28 Jan 2025 Shift 1

Solution:

Let the general point on the line be $Q(2\lambda + 1, \lambda + 2, 3\lambda + 1)$.
The direction ratios of the line are $\vec{d} = (2, 1, 3)$.
Let $Q$ be the foot of the perpendicular from $P(4, 4, 3)$ to the line.
Vector $\vec{PQ} = (2\lambda + 1 – 4, \lambda + 2 – 4, 3\lambda + 1 – 3) = (2\lambda – 3, \lambda – 2, 3\lambda – 2)$.

Since $\vec{PQ} \perp \vec{d}$:
$2(2\lambda – 3) + 1(\lambda – 2) + 3(3\lambda – 2) = 0$
$4\lambda – 6 + \lambda – 2 + 9\lambda – 6 = 0$
$14\lambda – 14 = 0 \Rightarrow \lambda = 1$.

Coordinates of foot of perpendicular $Q$:
$Q = (2(1)+1, 1+2, 3(1)+1) = (3, 3, 4)$.

Let the image be $R(\alpha, \beta, \gamma)$.
$Q$ is the midpoint of $PR$.
$\frac{\alpha + 4}{2} = 3 \Rightarrow \alpha + 4 = 6 \Rightarrow \alpha = 2$
$\frac{\beta + 4}{2} = 3 \Rightarrow \beta + 4 = 6 \Rightarrow \beta = 2$
$\frac{\gamma + 3}{2} = 4 \Rightarrow \gamma + 3 = 8 \Rightarrow \gamma = 5$

We need $\alpha + \beta + \gamma = 2 + 2 + 5 = 9$.

Ans. (1)