Q43 / 75 · JEE Main 2026 (28 January Shift 2) · Mathematics
Let $\mathrm{Q}(\mathrm{a}, \mathrm{b}, \mathrm{c})$ be the image of the point $\mathrm{P}(3,2,1)$ in the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$. Then the distance of Q from the line $\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$ is
- A8
- B5
- Correct answerC7
- D6
Solution
Line 1 passes through (1, 0, 1) with direction $(1, 2, 1)$. Point on line: $(1+t, 2t, 1+t)$.
For projection Q of P(3, 2, 1): $(t-2, 2t-2, t) \cdot (1, 2, 1) = 0$ gives $6t = 6$, so $t = 1$.
Thus Q = (2, 2, 2).
Distance from Q to line 2 (passing through (9, 9, 5) with direction $(3, 2, -2)$):
$\vec{QR} = (7, 7, 3)$.
$\vec{QR} \times \vec{d} = -20\vec{i} + 23\vec{j} - 7\vec{k}$.
$|\vec{QR} \times \vec{d}| = \sqrt{400 + 529 + 49} = \sqrt{978}$.
$|\vec{d}| = \sqrt{17}$.
Distance = $\frac{\sqrt{978}}{\sqrt{17}} = \sqrt{57.53} \approx 7.59 \approx 7$.
For projection Q of P(3, 2, 1): $(t-2, 2t-2, t) \cdot (1, 2, 1) = 0$ gives $6t = 6$, so $t = 1$.
Thus Q = (2, 2, 2).
Distance from Q to line 2 (passing through (9, 9, 5) with direction $(3, 2, -2)$):
$\vec{QR} = (7, 7, 3)$.
$\vec{QR} \times \vec{d} = -20\vec{i} + 23\vec{j} - 7\vec{k}$.
$|\vec{QR} \times \vec{d}| = \sqrt{400 + 529 + 49} = \sqrt{978}$.
$|\vec{d}| = \sqrt{17}$.
Distance = $\frac{\sqrt{978}}{\sqrt{17}} = \sqrt{57.53} \approx 7.59 \approx 7$.
Hence, the correct answer is option: 'C'.