JEE Main 2025 — Vector Algebra Question with Solution

$\begin{aligned} & \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{array}\right|=\hat{i}(-7)+7 \hat{j}+7 \hat{k} \\ & \hat{a}= \pm \frac{(-7 \hat{i}+7 \hat{j}+7 \hat{k})}{\sqrt{7^2+7^2+7^2}}= \pm\left(\frac{-\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}\right) \end{aligned}$ $\begin{aligned} & \text { Now, } \cos \theta= \pm \frac{(-1+1+1)}{\sqrt{3} \cdot \sqrt{3}}= \pm \frac{1}{3} \\ & \Rightarrow \cos ^{-1}\left(\frac{-1}{3}\right) \Rightarrow \hat{a}=\frac{-(-\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}} \\ & \hat{a}=\frac{\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}} \end{aligned}$ $\begin{aligned} & \cos \frac{\pi}{3}=\frac{1-\alpha-1}{\sqrt{3} \cdot \sqrt{\alpha^2+2}} \\ & \frac{1}{2}=\frac{-\alpha}{\sqrt{3} \cdot \sqrt{\alpha^2+2}} \rightarrow \alpha < 0 \\ & 3\left(\alpha^2+2\right)=4 \alpha^2 \\ & 6=\alpha^2 \\ & \alpha= \pm \sqrt{6} \end{aligned}$ Clearly, $\alpha =-\sqrt{6}$