Three Dimensional Geometry – The Plane – JEE Main 22 Jan 2026 Shift 1

Solution:

Let the given line be $L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$.

Since $Q(5, b, c)$ is the image of $P(1, 2, a)$ in line $L$, the midpoint $M$ of segment $PQ$ must lie on the line $L$.

Midpoint $M = \left(\frac{1+5}{2}, \frac{2+b}{2}, \frac{a+c}{2}\right) = \left(3, \frac{b+2}{2}, \frac{a+c}{2}\right)$.

Substitute $M$ into the equation of the line:
$$ \frac{3-6}{3} = \frac{\frac{b+2}{2}-7}{2} = \frac{\frac{a+c}{2}-7}{-2} $$
$$ -1 = \frac{b+2-14}{4} = \frac{a+c-14}{-4} $$

From the first equality:
$$ -1 = \frac{b-12}{4} \Rightarrow b – 12 = -4 \Rightarrow b = 8 $$

From the second equality:
$$ -1 = \frac{a+c-14}{-4} \Rightarrow a + c – 14 = 4 \Rightarrow a + c = 18 \quad \dots(1) $$

Also, the line segment $PQ$ is perpendicular to the line $L$. The direction ratios of $PQ$ are $(5-1, b-2, c-a) = (4, 6, c-a)$ (since $b=8$).
The direction ratios of line $L$ are $(3, 2, -2)$.

Dot product must be zero:
$$ 3(4) + 2(6) – 2(c-a) = 0 $$
$$ 12 + 12 – 2(c-a) = 0 $$
$$ 24 = 2(c-a) \Rightarrow c – a = 12 \quad \dots(2) $$

Solving (1) and (2):
Adding the equations: $(a+c) + (c-a) = 18 + 12 \Rightarrow 2c = 30 \Rightarrow c = 15$.
Substituting $c=15$ into (1): $a + 15 = 18 \Rightarrow a = 3$.

Now, find $a^2 + b^2 + c^2$:
$$ a^2 + b^2 + c^2 = (3)^2 + (8)^2 + (15)^2 $$
$$ = 9 + 64 + 225 $$
$$ = 298 $$

Ans. (4)