Find the area of the parallelogram whose adjacent sides are given by the vectors $(3,1,4)$ and $(1,-1,1)$.

Let the two adjacent side vectors be $\vec{a}=3\hat{i}+\hat{j}+4\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$.The area of a parallelogram is given by the magnitude of the cross product of its adjacent sides:
$$\text{Area}=|\vec{a}\times \vec{b}|$$
First, find $\vec{a}\times \vec{b}$:
$$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& 4\\ 1& -1& 1\end{array}\right|$$
$$=\hat{i}(1-(-4))-\hat{j}(3-4)+\hat{k}(-3-1)$$
$$=5\hat{i}+\hat{j}-4\hat{k}$$
Now, find the magnitude:
$$|\vec{a}\times \vec{b}|=\sqrt{5^2+1^2+(-4{)}^2}=\sqrt{25+1+16}=\sqrt{42}$$
The area is $\sqrt{42}$ square units.