Reply To: Question 14 — JEE Main 28 Jan 2026 (S I)

shahkb4
3 June, 2026 at 8:15 pm #1056

Solution

Point S which is equidistant from lines PQ and PR , lies on angle bisector of $\angle \mathrm{QPR}$
$|\overrightarrow{\mathrm{PR}}|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+16}=9$
$\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}+16=81$
$\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=65$
$|\overrightarrow{\mathrm{PQ}}|=\sqrt{4+1+4}=3$
S divides QR in $1: 3$ ratio


$\mathrm{S}\left(\frac{\mathrm{a}-6}{4}, \frac{\mathrm{~b}-3}{4}, \frac{1}{2}\right)$
$\overrightarrow{\mathrm{PS}}=\frac{(\mathrm{a}-6) \hat{\mathrm{i}}}{4}+\frac{(\mathrm{b}-3) \hat{\mathrm{j}}}{4}+\frac{\hat{\mathrm{k}}}{2}$
$\overrightarrow{\mathrm{PS}}=\frac{\hat{\imath}-7 \hat{\mathrm{j}}+2 \hat{k}}{4}=\frac{(\mathrm{a}-6) \hat{\mathrm{i}}+(\mathrm{b}-3) \hat{\mathrm{j}}+2 \hat{k}}{4}$
We get $a=7, b=-4$
$3 a-4 b=3(7)-4(-4)=21+16=37$ Answer(37)