Given : $\overrightarrow {a} = 3 \hat{i} - \hat {j} + 2 \hat {k} ; \overrightarrow {b} = \hat {i} - \hat{j} -3 \hat{k} ; \overrightarrow {c} = 4 \hat {i} -3\hat{j} + \hat{k}$
Area of $\triangle ABC = \frac{1}{2} |\overrightarrow {AB} \times \overrightarrow {AC}|$
$\begin{align*}\overrightarrow {AB} = \overrightarrow {b} - \overrightarrow {a} &= (\hat {i} - \hat{j} -3\hat{k}) - (3 \hat{i} -\hat{j} +2 \hat {j}) \\ & = -2 \hat{i} -5 \hat {k} \end{align*}$
$\begin{align*} \overrightarrow {AC} = \overrightarrow {c} - \overrightarrow {a} & = (4 \hat{i} -3 \hat{j} +\hat{k}) - (3 \hat{i} -\hat{j} +2 \hat{k}) \\ & = \hat{i} -2\hat{j} -\hat{k} \end{align*}$
$\overrightarrow {AB} \times \overrightarrow {AC} = \begin{vmatrix} \hat{i} & \hat {j} & \hat{k} \\ -2 & 0 & -5 \\ 1 & -2 & -1 \end{vmatrix} = -10 \hat{i}-7 \hat{j} +4\hat{k}$
$|\overrightarrow {AB} \times \overrightarrow {AC} | = \sqrt{(-10)^2 + (-7) ^2 + (4)^2} = \sqrt{165}$
Hence the area of the triangle is
$\frac{1}{2}|\overrightarrow {AB} \times \overrightarrow {AC}| = \frac{1}{2} \sqrt {165} sq.units$