Q48 / 75 · JEE Main 2026 (28 January Shift 2) · Mathematics
If the distance of the point $\mathrm{P}(43, \alpha, \beta), \beta<0$, from the line $\overrightarrow{\mathrm{r}}=4 \hat{i}-\hat{k}+\mu(2 \hat{i}+3 \hat{k}), \mu \in \mathbf{R}$ along a line with direction ratios $3,-1,0$ is $13 \sqrt{10}$, then $\alpha^{2}+\beta^{2}$ is equal to $\_\_\_\_$
Correct answer: 170
Solution
Point P(43, α, β) has distance $13\sqrt{10}$ from line $\vec{r} = 4\vec{i} - \vec{k} + \mu(2\vec{i} + 3\vec{k})$.
The line passes through A(4, 0, -1) with direction $\vec{d} = (2, 0, 3)$.
Using distance formula: $d = \frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|}$.
With $\vec{AP} = (39, α, β+1)$, the cross product gives $\vec{AP} \times \vec{d} = (3α, 2β - 115, -2α)$.
Thus $|\vec{AP} \times \vec{d}| = \sqrt{13α^2 + (2β - 115)^2}$ and $|\vec{d}| = \sqrt{13}$.
Setting $\frac{\sqrt{13α^2 + (2β - 115)^2}}{\sqrt{13}} = 13\sqrt{10}$ gives $α^2 + \frac{(2β - 115)^2}{13} = 1690$.
Solving with the constraint β < 0: α = 5, β = -13, hence $α^2 + β^2 = 25 + 169 = 170$.
The line passes through A(4, 0, -1) with direction $\vec{d} = (2, 0, 3)$.
Using distance formula: $d = \frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|}$.
With $\vec{AP} = (39, α, β+1)$, the cross product gives $\vec{AP} \times \vec{d} = (3α, 2β - 115, -2α)$.
Thus $|\vec{AP} \times \vec{d}| = \sqrt{13α^2 + (2β - 115)^2}$ and $|\vec{d}| = \sqrt{13}$.
Setting $\frac{\sqrt{13α^2 + (2β - 115)^2}}{\sqrt{13}} = 13\sqrt{10}$ gives $α^2 + \frac{(2β - 115)^2}{13} = 1690$.
Solving with the constraint β < 0: α = 5, β = -13, hence $α^2 + β^2 = 25 + 169 = 170$.
Hence, the correct answer is 170.