3. Given that the equation $x^{2}+(a-6) x+a=0$ has two integer roots with respect to $x$. Find the value of $a$.

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(Tip: Let the two integer roots be $x_{1} \geqslant x_{2}$, by Vieta's formulas we get $x_{1}+x_{2}=6-a, x_{1} x_{2}=a$, eliminating $a$, we get $x_{1} x_{2}+x_{1}+x_{2}$ $=6,\left(x_{1}+1\right)\left(x_{2}+1\right)=7$. Solving, we get $x_{1}=6, x_{2}=0$ or $x_{1}=$ $-2, x_{2}=-8$. Therefore, $a=x_{1} x_{2}=0$ or 16.)