Answer
$a\sqrt {c^{2}+b^{2}}$
Work Step by Step
The area of the parallelogram is given as:
$||⟨a,0,0⟩\times ⟨0,b,c⟩||$
Thus we have:
$⟨a,0,0⟩\times ⟨0,b,c⟩=\begin{vmatrix}\textbf{i}&\textbf{j}&\textbf{k}\\a&0&0\\0&b&c\end{vmatrix}$
$=\textbf{i}(0-0)-\textbf{j}(ac-0)+\textbf{k}(ab-0)$
$=-ac\textbf{j}+ab\textbf{k}$
$||⟨a,0,0⟩\times ⟨0,b,c⟩||=$
$\sqrt {(-ac)^{2}+(ab)^{2}}=\sqrt {a^{2}c^{2}+a^{2}b^{2}}$
$=\sqrt {a^{2}(c^{2}+b^{2})}=a\sqrt {c^{2}+b^{2}}$
