[shahkb4]

shahkb4

3 June, 2026 at 8:16 pm #1087

If the distance of the point $\mathrm{P}(43, \alpha, \beta), \beta<0$, from the
line $\vec{r}=4 \hat{i}-\hat{k}+\mu(2 \hat{i}+3 \hat{k}), \mu \in R$ along a line with direction ratios $3,-1,0$ is $13 \sqrt{10}$, then $\alpha^{2}+\beta^{2}$ is equal to $\_\_\_\_$

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[shahkb4]

shahkb4

3 June, 2026 at 8:16 pm #1088

Solution

line $\mathrm{L}_{1}: \frac{\mathrm{x}-4}{2}=\frac{\mathrm{y}-0}{0}=\frac{\mathrm{z}+1}{3}=\lambda$
line $\mathrm{L}_{2}: \frac{\mathrm{x}-43}{3}=\frac{\mathrm{y}-\alpha}{-1}=\frac{\mathrm{z}-\beta}{0}=\mu$
$\mathrm{Q}(2 \lambda+4,0,3 \lambda-1) \equiv \mathrm{Q}(3 \mu+43,-\mu+\alpha, \beta)$
$2 \lambda+4=3 \mu+43 \Rightarrow 2 \lambda=3 \mu+39$
$0=-\mu+\alpha=\alpha=\mu$

\begin{figure}
\captionsetup{labelformat=empty}
\caption{[JEE MAIN 28 Jan 2026 S II]}
\includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/27a95272-36b0-4ba6-b451-34902f9c19d8-17.jpg?height=351&width=530&top_left_y=339&top_left_x=910}
\end{figure}
$3 \lambda-1=\beta$
$\mathrm{Q}(3 \alpha+43,0, \beta) \mathrm{P}(43, \alpha, \beta)$
$\mathrm{PQ}=13 \sqrt{10} \Rightarrow \sqrt{9 \alpha^{2}+\alpha^{2}+0}=13 \sqrt{10}$
$\Rightarrow|\alpha| \sqrt{10}=13 \sqrt{10} \Rightarrow \alpha= \pm 13$
Case I : $\alpha=13, \mu=13$, given $\beta<0$
From (1), $2 \lambda=3 \mu+39=78 \Rightarrow \lambda=39$
From (2), $\beta=3 \lambda-1=3 \times 39-1>0$ (rejected)
Case II : $\alpha=-13, \mu=-13$
From (1), $2 \lambda=-39+39 \Rightarrow \lambda=0$
From (3), $\beta=3 \lambda-1=-1$
Now $\alpha^{2}+\beta^{2}=169+1=170$ Answer(170)

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