JEE Main 2024 — Three Dimensional Geometry Question with Solution

Given: 

${\mathrm{L}}_1\equiv \frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}=\mathrm{\lambda }$

$\Rightarrow \mathrm{M}(\mathrm{\lambda },1+2\mathrm{\lambda },2+3\mathrm{\lambda })$

$\Rightarrow \overrightarrow{\mathrm{PM}}=\left(\mathrm{\lambda }-1\right)\hat{\mathrm{i}}+\left(1+2\mathrm{\lambda }\right)\hat{\mathrm{j}}+\left(3\mathrm{\lambda }-5\right)\hat{\mathrm{k}}$

So, $\overrightarrow{\mathrm{PM}}$ is perpendicular to line ${\mathrm{L}}_1$

$\Rightarrow \overrightarrow{\mathrm{PM}}·\overrightarrow{\mathrm{b}}=0$

$\Rightarrow \overrightarrow{\mathrm{PM}}.\overrightarrow{\mathrm{b}}=\left[\left(\mathrm{\lambda }-1\right)\hat{\mathrm{i}}+\left(1+2\mathrm{\lambda }\right)\hat{\mathrm{j}}+\left(3\mathrm{\lambda }-5\right)\hat{\mathrm{k}}\right].\left(\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)$

$\Rightarrow \mathrm{\lambda }-1+4\mathrm{\lambda }+2+9\mathrm{\lambda }-15=0$

$\Rightarrow 14\mathrm{\lambda }=14$

$\Rightarrow \mathrm{\lambda }=1$

$\Rightarrow \mathrm{M}=(1,3,5)$

Now, $\mathrm{M}$ is midpoint of $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$

$\Rightarrow \left(1,3,5\right)\equiv \left(\frac{1+\mathrm{\alpha }}{2},\frac{0+\mathrm{\beta }}{2},\frac{7+\mathrm{\gamma }}{2}\right)$

$\Rightarrow \frac{1+\mathrm{\alpha }}{2}=1, \frac{\mathrm{\beta }}{2}=3, \frac{7+\mathrm{\gamma }}{2}=5$

$\Rightarrow \mathrm{\alpha }=1, \mathrm{\beta }=6, \mathrm{\gamma }=3$

Now, required line having direction cosine $\left(\mathrm{l},\mathrm{m},\mathrm{n}\right)$, where given $m=\cos \left(\frac{-2\mathrm{\pi }}{3}\right)=\frac{-1}{2} \& n=\cos \left(\frac{3\mathrm{\pi }}{4}\right)=\frac{-1}{\sqrt{2}}$

Also, ${\mathrm{l}}^2+{\mathrm{m}}^2+{\mathrm{n}}^2=1$

$\Rightarrow {\mathrm{l}}^2+{\left(-\frac{1}{2}\right)}^2+{\left(-\frac{1}{\sqrt{2}}\right)}^2=1$

$\Rightarrow {\mathrm{l}}^2=\frac{1}{4}$

$\Rightarrow \mathrm{l}=\frac{1}{2}$

[Line make acute angle with $\mathrm{x}-$axis]

So, the equation of line passing through $\left(1,6,3\right)$ will be given by,

$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+6\hat{\mathrm{j}}+3\hat{\mathrm{k}})+\mathrm{\mu }\left(\frac{1}{2}\hat{\mathrm{i}}-\frac{1}{2}\hat{\mathrm{j}}-\frac{1}{\sqrt{2}}\hat{\mathrm{k}}\right)$

Option $\left(3\right)$ $\left(3,4,3-2\sqrt{2}\right)$ satisfying for $\mathrm{\mu }=4$