JEE MAIN - Mathematics (2024 - 6th April Morning Shift - No. 5)
Explanation
$$A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1), D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$$ are vertices of a quadrilateral
$$\begin{aligned} & \overrightarrow{A C}=(2 \hat{i}+2 \hat{j}+\hat{k})-(3 \hat{i}+\hat{j}-\hat{k}) \\ & =-\hat{i}+\hat{j}+2 \hat{k} \\ & \overrightarrow{B D}=\left(\frac{10}{3} \hat{i}+\frac{2}{3} \hat{j}-\frac{1}{3} \hat{k}\right)-\left(\frac{5}{3} \hat{i}+\frac{7}{3} \hat{j}+\frac{1}{3} \hat{k}\right) \\ & \overrightarrow{B D}=\frac{5}{3} \hat{i}-\frac{5}{3} \hat{j}-\frac{2}{3} \hat{k} \\ & \text { Area }=\frac{1}{2}|\overrightarrow{A C} \times \overrightarrow{B D}| \\ & =\frac{1}{2}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ \frac{5}{3} & \frac{-5}{3} & \frac{-2}{3} \end{array}\right| \\ & =\frac{1}{2} \sqrt{\left(\frac{8}{3}\right)^2+\left(\frac{8}{3}\right)^2}\left[\because \overrightarrow{A C} \times \overrightarrow{B D}=\frac{8}{3} \hat{i}+\frac{8}{3} \hat{j}\right] \\ & =\frac{1}{2} \times \frac{8}{3} \times \sqrt{2}=\frac{4 \sqrt{2}}{3} \\ \end{aligned}$$
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