3. Given that $n$ is an integer, and the quadratic equation in $x$
$$
(n-1)^{2} x^{2}-5 n(n-1) x+\left(6 n^{2}-n-1\right)=0
$$

has at least one integer root. Then the sum of all possible values of $n$ is $\qquad$
$$
[(n-1) x-(2 n-1)][(n-1) x-(3 n+1)]=0 .
$$

Since $n \neq 1$, then
$$
\begin{array}{l}
x_{1}=\frac{2 n-1}{n-1}=2+\frac{1}{n-1}, \\
x_{2}=\frac{3 n+1}{n-1}=3+\frac{4}{n-1} .
\end{array}
$$

Because the original equation has at least one integer root, and $n$ is an integer, so, $n-1= \pm 1, \pm 2, \pm 4$.
Solving gives $n=2,0,3,-1,5,-3$.
Therefore, the sum of all $n$ values is 6.