Find $\lambda$ if volume of parallelopiped having edges $\overline{i}+\overline{j}, 3\overline{i}-\overline{j}$ and $3\overline{j}+\lambda \overline{k}$ is 16 cubic units.

Given edges of parallelepiped are 

$\overline{a}=\overline{i}+\overline{j},\overline{b}=3\overline{i}-\overline{j},\overline{c}=3\overline{j}+\lambda \overline{k}$

Given volume of the parallelepiped $=16\text{ cubic units. }$

$\Rightarrow \left|\left[\begin{array}{l}\overline{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overline{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overline{c}\end{array}\right]\right|=16$

Now $\left[\begin{array}{l}\overline{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overline{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overline{c}\end{array}\right]=\left|\begin{array}{ccc}1& 1& 0\\ 3& -1& 0\\ 0& 3& \lambda \end{array}\right|=1(-\lambda )-1\left(3\lambda \right)+0$

$=-\lambda -3\lambda =-4\lambda$

$\therefore |-4\lambda |=16\Rightarrow \left|4\lambda \right|=16\Rightarrow 4\lambda =\pm 16\Rightarrow \boxed{\lambda =\pm 4}$