JEE Main 2026 Jan 22 Shift 1, Mathematics Q11: Lines in Space

JEE Main2026Jan 22, Shift 1Mathematics

If the image of the point $P (1, 2, a)$ in the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ is $Q (5, b, c)$ , then $a^{2} + b^{2} + c^{2}$ is equal to

A
293
B
264
C
298
D
283

Solution

Rewrite the line: $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ .
Direction $d = (3, 2, - 2)$ , point on line $L_{0} = (6, 7, 7)$ .

Midpoint $M$ of $P Q = (3, \frac{b + 2}{2}, \frac{c + a}{2})$ must lie on the line:

$\frac{3 - 6}{3} = \frac{( b + 2 ) /2 - 7}{2} = \frac{( c + a ) /2 - 7}{- 2}$ , i.e. $- 1 = \frac{b - 12}{4} = \frac{c + a - 14}{- 4}$ .

So $b - 12 = - 4 \Rightarrow b = 8$ , and $c + a = 18$ …(i).

$P Q = (4, b - 2, c - a)$ must be perpendicular to $d$ : $(4, b - 2, c - a) \cdot (3, 2, - 2) = 0$ , i.e. $12 + 2 (b - 2) - 2 (c - a) = 0$ . With $b = 8$ : $c - a = 12$ …(ii).

Adding (i) and (ii): $2 c = 30 \Rightarrow c = 15$ , so $a = 3$ .

$a^{2} + b^{2} + c^{2} = 9 + 64 + 225 = 298$ .

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JEE Main 2026 Jan 22 Shift 1, Mathematics Q11: Lines in Space

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