JEE 2023 — Mathematics PYQ

Solution:

Vector relation: $\vec{r}\times \vec{b}-\vec{c}\times \vec{b}=0⟹(\vec{r}-\vec{c})\times \vec{b}=0$. Iska matlab $\vec{r}-\vec{c}$ aur $\vec{b}$ parallel hain.

$\vec{r}$ ki form: $\vec{r}=\vec{c}+\lambda \vec{b}=(5+\lambda )\hat{i}+(-3-\lambda )\hat{j}+(3+2\lambda )\hat{k}$.

Condition $\vec{r}\cdot \vec{a}=0$:

$(5+\lambda )(1)+(-3-\lambda )(2)+(3+2\lambda )(3)=0$

$5+\lambda -6-2\lambda +9+6\lambda =0⟹5\lambda +8=0⟹\lambda =-\frac{8}{5}$.

Vector $\vec{r}$: $\vec{r}=\frac{17}{5}\hat{i}-\frac{7}{5}\hat{j}-\frac{1}{5}\hat{k}$.

Final Calculation: $25|\vec{r}{|}^2=25\left(\frac{17^2+7^2+1^2}{25}\right)=289+49+1=339$.

Correct Option: (3)