QJEEM24Mace2b: JEE Main 2024 Mathematics - Class 12 JEEM24Mace2b Solution

$A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1), D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$ are vertices of a quadrilateral

$\begin{aligned}& \overrightarrow{A C}=(2 \hat{i}+2 \hat{j}+\hat{k})-(3 \hat{i}+\hat{j}-\hat{k}) \\& =-\hat{i}+\hat{j}+2 \hat{k} \\& \overrightarrow{B D}=\left(\frac{10}{3} \hat{i}+\frac{2}{3} \hat{j}-\frac{1}{3} \hat{k}\right)-\left(\frac{5}{3} \hat{i}+\frac{7}{3} \hat{j}+\frac{1}{3} \hat{k}\right) \\& \overrightarrow{B D}=\frac{5}{3} \hat{i}-\frac{5}{3} \hat{j}-\frac{2}{3} \hat{k} \\& \text { Area }=\frac{1}{2}|\overrightarrow{A C} \times \overrightarrow{B D}| \\& =\frac{1}{2}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\-1 & 1 & 2 \\\frac{5}{3} & \frac{-5}{3} & \frac{-2}{3}\end{array}\right| \\& =\frac{1}{2} \sqrt{\left(\frac{8}{3}\right)^2+\left(\frac{8}{3}\right)^2}\left[\because \overrightarrow{A C} \times \overrightarrow{B D}=\frac{8}{3} \hat{i}+\frac{8}{3} \hat{j}\right] \\& =\frac{1}{2} \times \frac{8}{3} \times \sqrt{2}=\frac{4 \sqrt{2}}{3} \\\end{aligned}$