JEE 2023 — Mathematics PYQ

Here $\overrightarrow{AC}=(-2-2)\hat{i}+(-3-1)\hat{j}+(5-1)\hat{k}$

$=-4\hat{i}-4\hat{j}+4\hat{k}$

$\overrightarrow{BD}=(1-1)\hat{i}+(-6-2)\hat{j}+(-7-5)\hat{k}$

$=-8\hat{j}-12\hat{k}$

So, area of quadrilateral $=\frac{1}{2}|\overrightarrow{AC}\times \overrightarrow{BD}|$

$=\frac{1}{2}\Vert \begin{array}{ccc}\hat{i}& amp;\hat{j}& amp;\hat{k}\\ -4& amp;-4& amp;4\\ 0& amp;-8& amp;-12\end{array}\Vert$

$=\frac{1}{2}|(48+32)\hat{i}-(48-0)\hat{j}+(32-0)\hat{k}|$

$=\frac{1}{2}|80\hat{i}-48\hat{j}+32\hat{k}|$

$=\frac{1}{2}\cdot 16|5\hat{i}-3\hat{j}+2\hat{k}|$

$=8\sqrt{25+9+4}=8\sqrt{38}$ sq units.