JEE Main 2026 Jan 21 Shift 2, Mathematics Q7: Lines in Space

Equal angles with positive axes ⇒ direction ratios $(1,1,1)$. So $L:\frac{x+3}{1}=\frac{y-5}{1}=\frac{z-2}{1}=\lambda$.

Let $P=(-2,r,1)$ and let $R$ be the foot of perpendicular from $P$ to $L$. $R=(\lambda -3,\lambda +5,\lambda +2)$, and $\overrightarrow{PR}=(\lambda -1,\lambda +5-r,\lambda +1)$.

$\overrightarrow{PR}\cdot \vec{d}=0$: $(\lambda -1)+(\lambda +5-r)+(\lambda +1)=0\Rightarrow 3\lambda -r+5=0\Rightarrow \lambda =\frac{r-5}{3}$.

So $R=\left(\frac{r-14}{3},\frac{r+10}{3},\frac{r+1}{3}\right)$ and $|PR{|}^2=\frac{14}{3}$:

${\left(\frac{r-14}{3}+2\right)}^2+{\left(\frac{r+10}{3}-r\right)}^2+{\left(\frac{r+1}{3}-1\right)}^2=\frac{14}{3}$

$\frac{(r-8{)}^2+(10-2r{)}^2+(r-2{)}^2}{9}=\frac{14}{3}$, i.e. $6r^2-60r+126=0$, or $r^2-10r+21=0$ ⇒ $r=3$ or $r=7$.

Sum $=10$.