A) \[5\sqrt{3}\]
B) \[6\sqrt{3}\]
C) \[\sqrt{26}\]
D) \[\sqrt{42}\]
Correct Answer: D
Solution :
Area of parallelogram \[=\frac{1}{2}|{{\mathbf{d}}_{1}}\times {{\mathbf{d}}_{2}}|\] \[{{\mathbf{d}}_{1}}\times {{\mathbf{d}}_{2}}=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & -2 \\ 1 & 3 & -4 \\ \end{matrix} \right|\]; \[{{\mathbf{d}}_{1}}\times {{\mathbf{d}}_{2}}=2\mathbf{i}+10\mathbf{j}+8\mathbf{k}\] \[\therefore \,\,\,\frac{1}{2}|{{\mathbf{d}}_{1}}\times {{\mathbf{d}}_{2}}|\,=\,\frac{1}{2}\sqrt{4+100+64}=\frac{1}{2}\sqrt{168}=\sqrt{42}\].
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