5. Given the equation $6 x^{2}+2(m-13) x+12-m$ $=0$ has exactly one positive integer solution. Then the value of the integer $m$ is
5. Since $\Delta=4(m-13)^{2}-24(12-m)$ is a perfect square, there exists a non-negative integer $y$ such that
$$
(m-13)^{2}-6(12-m)=y^{2} \text {, }
$$

i.e., $(m-10-y)(m-10+y)=3$.
Since $m-10-y \leqslant m-10+y$, we have
$$
\left\{\begin{array} { l } 
{ m - 1 0 - y = 1 , } \\
{ m - 1 0 + y = 3 }
\end{array} \text { or } \left\{\begin{array}{l}
m-10-y=-3, \\
m-10+y=-1 .
\end{array}\right.\right.
$$

Solving these, we get $m=12$ or $m=8$.
If $m=12$, then the original equation has no positive integer solutions, which is a contradiction.
If $m=8$, then the original equation yields $x=1$ or $x=\frac{2}{3}$. This satisfies the conditions.
Therefore, the required $m=8$.