Step 1: Write the two vectors in component form.
$\vec{A} = (2, 3, -1)$ and $\vec{B} = (5, 0, -4)$, since $\vec{B}$ has no $\hat{j}$ part.
The parallelogram area formed by two adjacent side vectors is the magnitude of their cross product, $|\vec{A} \times \vec{B}|$.
Step 2: Compute each component using 2x2 minors.
$i$-component: drop column 1, take $\begin{vmatrix} 3 & -1 \\ 0 & -4 \end{vmatrix} = (3)(-4) - (-1)(0) = -12$.
$j$-component: drop column 2, take $-\begin{vmatrix} 2 & -1 \\ 5 & -4 \end{vmatrix} = -[(2)(-4)-(-1)(5)] = -(-3) = 3$.
$k$-component: drop column 3, take $\begin{vmatrix} 2 & 3 \\ 5 & 0 \end{vmatrix} = (2)(0)-(3)(5) = -15$.
This gives $\vec{A} \times \vec{B} = -12\hat{i} + 3\hat{j} - 15\hat{k}$, the same result checked a second way.
Step 3: Take the magnitude of this vector.
$|\vec{A} \times \vec{B}| = \sqrt{144+9+225} = \sqrt{378}$.
Since $19.4^2 = 376.36$ and $19.5^2 = 380.25$, the root lies between them, close to $19.44$.
Checking, $19.44^2 = 377.99$, which confirms the value.
Final Answer:
The parallelogram has area
\[ \boxed{19.44} \]