The area (in sq. units) of the parallelogram whose diagonals are along the vectors

$8\widehat{i}-6\widehat{j}\text{ and }3\widehat{i}+4\widehat{j}-12\widehat{k}\text{, is }$

$\text{ Let }\overrightarrow{a}=8\widehat{i}-6\widehat{j}\text{ and }\overrightarrow{b}=3\widehat{i}+4\widehat{j}-12\widehat{k}$

Area of parallelogram, $A=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|$

$\begin{array}{l}\therefore \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 8& -6& 0\\ 3& 4& -12\end{array}\right|=72\widehat{i}-(-96)\widehat{j}+50\widehat{k}\\ \therefore |\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{5184+9216}+2500=\sqrt{16900}=130\\ A=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|=\frac{1}{2}\times 130=65\end{array}$