Given that $\overrightarrow a=3\hat i+\hat j-2\hat k$ and
$\overrightarrow b=\hat i-3\hat j+4\hat k$ are diagonals of the parallelogram.
We know that$\overrightarrow a\times\overrightarrow b=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{array}\right|$
$=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\3&1&-2\\1&-3&4\end{array}\right|$ $=\:\hat i(4-6)-\hat j(12-(-2))+\hat k(-9-1)$
$=-2\hat i-14\hat j-10\hat k$
$|\overrightarrow a\times\overrightarrow b|=\sqrt{(-2)^2+(-14)^2+(-10)^2}$
$=\sqrt{4+196+100}=\sqrt{300}=10\sqrt3$
We know that area of the parallelogram =$\large\frac{1}{2}|$$\overrightarrow a\times\overrightarrow b|$ where $\overrightarrow a\:\:and\:\:\overrightarrow b$ are diagonals of the prarallelogram.
Area of the parallelogram $=\large\frac{1}{2}$$10\sqrt3=5\sqrt3 \approx 8.66$