Reply To: Question 13 — JEE Main 28 Jan 2026 (S I)

Solution

$\because|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{c}}|=1$
As given $|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$
$\Rightarrow|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}-2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+|\overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{c}}|^{2}-2 \overrightarrow{\mathrm{~b}} \cdot \overrightarrow{\mathrm{c}}+|\overrightarrow{\mathrm{c}}|^{2}+|\overrightarrow{\mathrm{a}}|^{2}-2 \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=9$
$\Rightarrow 1+1-2 \vec{a} \cdot \vec{b}+1+1-2 \vec{b} \cdot \vec{c}+1+1-2 \vec{c} \cdot \vec{a}=9$
$\Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=-\frac{3}{2}$
$\because|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$=1+1+1+2\left(-\frac{3}{2}\right)=0$
$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0$
$|2 \overrightarrow{\mathrm{a}}+\mathrm{k}(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=3$
$\Rightarrow|2 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{a}} \mathrm{k}|=3 \Rightarrow|\overrightarrow{\mathrm{a}} \|(2-\mathrm{k})|=3$
$\Rightarrow 2-\mathrm{k} \mid=3 \Rightarrow 2-\mathrm{k}= \pm 3$
we get $\mathrm{k}=-1 ; \mathrm{k}=5$ Answer(4)