JEE Main 2026 — 21 January, Evening Shift — Question 7

Equation line is : $\frac{\mathrm{x}+3}{1}=\frac{\mathrm{y}-5}{1}=\frac{\mathrm{z}-2}{1}=\lambda$

∴ General point R on line is $\mathrm{R}(\lambda-3, \lambda+5, \lambda+2)$

$\overrightarrow{\mathrm{PR}} \equiv(\lambda-1, \lambda+5-\mathrm{r}, \lambda+1)$

Now $\overrightarrow{\mathrm{PR}} \cdot \overrightarrow{\mathrm{d}}=0$

$\Rightarrow(\lambda-1) 1+(\lambda+5-\mathrm{r}) 1+(\lambda+1) 1=0$

$\Rightarrow 3 \lambda-\mathrm{r}+5=0$

$\Rightarrow \lambda=\frac{\mathrm{r}-5}{3}$

$\therefore \mathrm{R} \equiv\left(\frac{\mathrm{r}-5}{3}-3, \frac{\mathrm{r}-5}{3}+5, \frac{\mathrm{r}-5}{3}+2\right)$

$\mathrm{R} \equiv\left(\frac{\mathrm{r}-14}{3}, \frac{\mathrm{r}+10}{3}, \frac{\mathrm{r}+1}{3}\right)$

Now $\mathrm{PR}=\sqrt{\frac{14}{3}} \Rightarrow(\mathrm{PR})^{2}=\frac{14}{3}$

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

$\Rightarrow \frac{(\mathrm{r}-8)^{2}}{9}+\frac{(10-2 \mathrm{r})^{2}}{9}+\frac{(\mathrm{r}-2)^{2}}{9}=\frac{14}{3}$

$\Rightarrow\left(\mathrm{r}^{2}-16 \mathrm{r}+64\right)+\left(100+4 \mathrm{r}^{2}-40 \mathrm{r}\right)+\left(\mathrm{r}^{2}-4 \mathrm{r}+4\right)=42$

$\Rightarrow 6 \mathrm{r}^{2}-60 \mathrm{r}+126=0$

$\Rightarrow \mathrm{r}^{2}-10 \mathrm{r}+21=0$

$\Rightarrow \mathrm{r}=3,7$

sum of possible value of $r$ is $=10$