$A(1, 2, -1)$, $B(2, 5, -2)$ and $C(4, 4, -3)$ are three vertices of a rectangle. What is the area of the rectangle?

The given vertices of the rectangle are $A(1, 2, -1)$, $B(2, 5, -2)$ and $C(4, 4, -3)$.

The direction ratios of the side $AB$ are $(2-1, 5-2, -2-(-1)) = (1, 3, -1)$.
The length of the side $AB$ is $\sqrt{1^2 + 3^2 + (-1)^2} = \sqrt{1 + 9 + 1} = \sqrt{11}$.

The direction ratios of the side $BC$ are $(4-2, 4-5, -3-(-2)) = (2, -1, -1)$.
The length of the side $BC$ is $\sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.

Now, checking the dot product of vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$:
$\overrightarrow{AB} \cdot \overrightarrow{BC} = (1)(2) + (3)(-1) + (-1)(-1) = 2 - 3 + 1 = 0$

Since the dot product is zero, $AB$ is perpendicular to $BC$, which means $A$, $B$, and $C$ are consecutive vertices forming a right angle at $B$.

The area of the rectangle is the product of its adjacent sides:
$\text{Area} = AB \times BC = \sqrt{11} \times \sqrt{6} = \sqrt{66} \text{ square units}$

Answer: $\sqrt{66}$ square units