Three real numbers $a$, $b$, and $c$ between $0$ and $1$ are chosen independently and at random. What is the probability that $a + 2b + 3c > 5$?
1. We start by considering the unit cube in the 3-dimensional space where each of the coordinates \(a\), \(b\), and \(c\) ranges from 0 to 1. This cube represents all possible combinations of \(a\), \(b\), and \(c\).

2. We need to find the region within this cube where the inequality \(a + 2b + 3c > 5\) holds. This inequality represents a plane in the 3-dimensional space.

3. To understand the region better, we first find the intercepts of the plane \(a + 2b + 3c = 5\) with the coordinate axes:
   - When \(b = 0\) and \(c = 0\), \(a = 5\). However, since \(a\) is constrained to be between 0 and 1, this intercept is outside the unit cube.
   - When \(a = 0\) and \(c = 0\), \(2b = 5\) or \(b = \frac{5}{2}\). This is also outside the unit cube.
   - When \(a = 0\) and \(b = 0\), \(3c = 5\) or \(c = \frac{5}{3}\). This is also outside the unit cube.

4. Since all intercepts are outside the unit cube, the plane \(a + 2b + 3c = 5\) does not intersect the unit cube. Therefore, the region where \(a + 2b + 3c > 5\) lies entirely outside the unit cube.

5. As a result, there are no points \((a, b, c)\) within the unit cube that satisfy the inequality \(a + 2b + 3c > 5\).

6. Therefore, the volume of the region within the unit cube where \(a + 2b + 3c > 5\) is zero.

The final answer is \(\boxed{0}\).