JEE Main 2024 — 8 April, Shift 1 — Question 28

$\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}$

$\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{c}}=17 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\vec{b}+\vec{c}=20 \hat{i}+5 \hat{j}-12 \hat{k}$

$\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}=-14 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}-14 \hat{\mathrm{k}}$ $(\overrightarrow{\mathrm{r}}-(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})) \times \overrightarrow{\mathrm{a}}=0$

$\mathrm{r}-(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\lambda \overrightarrow{\mathrm{a}}$ $\overrightarrow{\mathrm{r}}=\lambda \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}$

But $\vec{r} .(\vec{b}-\vec{c})=0$

$\Rightarrow(\lambda \vec{a}+\vec{b}+\vec{c}) \cdot(\vec{b}-\vec{c})=0$

$\Rightarrow \lambda \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{b}-\lambda \vec{a} \cdot \vec{c}-\vec{b} \cdot \vec{c}-\vec{c} \cdot \vec{c}=0$

$\lambda=\frac{\vec{c} \cdot \vec{c}-\vec{b} \cdot \vec{b}}{\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c}}=\frac{294-227}{-389=204}=\frac{-67}{593}$

$\therefore \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}-\frac{67}{593} \vec{a}$

$\Rightarrow 593 \overrightarrow{\mathrm{r}}+67 \overrightarrow{\mathrm{a}}=593(\overrightarrow{\mathrm{~b}}+\overrightarrow{\mathrm{c}})$ $\Rightarrow|\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|^{2}=569$