Q34 / 75 · JEE Main 2026 (22 January Shift 1) · Mathematics
If the image of the point $\mathrm{P}(1,2, a)$ in the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-\mathrm{z}}{2}$ is $\mathrm{Q}(5, b, \mathrm{c})$, then $a^{2}+b^{2}+c^{2}$ is equal to
- Correct answerA298
- B293
- C264
- D283
Solution
Line: point $(6,7,7)$, direction $(3,2,-2)$.
General point on line: $(6+3t, 7+2t, 7-2t)$.
Foot of perpendicular from $P(1,2,a)$: $\vec{PM} \cdot (3,2,-2) = 0$.
$3(5+3t)+2(5+2t)-2(7-2t-a) = 0$
$\Rightarrow 17t+11+2a = 0$.
Image $Q = 2M - P = (11+6t, 12+4t, 14-4t-a)$.
Given $Q_x = 5$: $11+6t = 5 \Rightarrow t = -1$.
From $17(-1)+11+2a = 0 \Rightarrow a = 3$.
$b = 12+4(-1) = 8$, $c = 14-4(-1)-3 = 15$.
$a^2+b^2+c^2 = 9+64+225 = 298$.
General point on line: $(6+3t, 7+2t, 7-2t)$.
Foot of perpendicular from $P(1,2,a)$: $\vec{PM} \cdot (3,2,-2) = 0$.
$3(5+3t)+2(5+2t)-2(7-2t-a) = 0$
$\Rightarrow 17t+11+2a = 0$.
Image $Q = 2M - P = (11+6t, 12+4t, 14-4t-a)$.
Given $Q_x = 5$: $11+6t = 5 \Rightarrow t = -1$.
From $17(-1)+11+2a = 0 \Rightarrow a = 3$.
$b = 12+4(-1) = 8$, $c = 14-4(-1)-3 = 15$.
$a^2+b^2+c^2 = 9+64+225 = 298$.
Hence, the correct answer is option: 'A'.