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# The area of the parallelogram having a diagonal $\overrightarrow{3i}+\overrightarrow{j}-\overrightarrow{k}$ and a side $\overrightarrow{i}-\overrightarrow{3j}+\overrightarrow{4k}$ is

$\begin{array}{1 1}(1)\;10\sqrt{3} &(2)\;6\sqrt{30}\\(3)\;\frac{3}{2}\sqrt{30}&(4)\;3\sqrt{30}\end{array}$

Let $\overrightarrow{AB}=\overrightarrow{i}-3\overrightarrow{j}+4 \overrightarrow{k}$
$\overrightarrow{AC}=3\overrightarrow{i} + \overrightarrow{j} - \overrightarrow{k}$
The vector area of the triangle ABC with adjacent sides $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is
$\qquad= \large\frac{1}{2}$$\overrightarrow{AB} \times \overrightarrow{AC} \qquad=\large \frac{1}{2}$$(\overrightarrow{i}-3 \overrightarrow{j} +4 \overrightarrow{k} ) \times 3(\overrightarrow{i} + \overrightarrow{j}- \overrightarrow{k})$
$\qquad= \large\frac{1}{2}$$\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\1 & -3 & 4 \\3 & 1 & -1 \end{vmatrix} \qquad= \large\frac{1}{2}$$ [ \overrightarrow{i} (3-4) -\overrightarrow{j}(-1-12)+\overrightarrow{k}(1+9)]$
$\qquad= \large\frac{1}{2}$$[- \overrightarrow{i} +13 \overrightarrow{j}+10 \overrightarrow{k}] Area of triangle = \large\frac{1}{2}$$ |- \overrightarrow{i} +13 \overrightarrow{j}+10 \overrightarrow{k}|$
$\qquad= \large\frac{1}{2} $$\sqrt {(-1)^2+13^2+10^2} \qquad= \large\frac{1}{2}$$ \sqrt {1+169+100}$
$\qquad= \large\frac{1}{2} $$\sqrt {270} \qquad= \large\frac{1}{2}$$ \sqrt {9 \times 30}$
$\qquad= \large\frac{3}{2} $$\sqrt {30} Area of parallelogram ABCD =2 \times area of \Delta ABC \qquad= 2 \times \large\frac{3}{2}$$\sqrt {30} =3 \sqrt {30}$
Hence 4 is the correct answer