This is really easy.we have this identity $\tan A +\tan B+\tan C=\tan A \tan B \tan C$ since $A+B+C=\pi$ so we have have to find the integral solutions of this $x+y+z=xyz$ where $x=\tan A,y=\tan B,z=\tan C$ It is not possible to find the solution when two of them $x,y,z$ are $1$.and $\triangle ABC$ can not be right angled.The triangle can not be obtused.Since if the triangle is obtused then one of them must be negative and other two will be positive.If we increase the value of the positive quantity then L.H.S will be greater than R.H.S.Hence $\triangle ABC$ must be acute.wlog asume $x \ge y \ge z \implies xyz=x+y+z \le 3x$ so, $yz \le 3$.so $y=2,z=1$.Hence the solution is $(\tan A,\tan B,\tan C)=(3,2,1).$