3. Given that the three sides of a triangle $a, b, c$ are unequal integers, and satisfy
$$
a b c+a b+b c+a c+a+b+c=119 .
$$

Then this triangle is a $\qquad$ triangle.
3. Right Angle.

The original equation can be transformed into $(a+1)(b+1)(c+1)=120$.
From the given information, $a+1$, $b+1$, and $c+1$ are also positive integers.
Assume $a < c$; the triangle can still exist. When $a+1$ takes the values $1, 2, 3$, $b$ and $c$ take positive integers that satisfy the conditions, but the triangle does not exist. For example, $120=3 \times 4 \times 10$, i.e., $a+1=3, b+1=4, c+1=10$, which means $a=2, b=3, c=9$. In this case, the triangle does not exist.

Only when $(a+1)(b+1)(c+1)=4 \times 5 \times 6$, i.e., $a=3, b=4, c=5$, does the triangle exist and is a right-angled triangle.