Clearly the equation gives $xabc < a$ i.e. $bc < \frac{1}{x}$.As $b,c$ are positive integers, $bc \geq 1$.So we have $\frac{1}{x} > 1$ i.e. $1 > x$.As $x$ is a integer this is equivalent to $x \leq 0$.But this is impossible as $x$ is positive.Hence there is no such $x$ This as it happens is completely wrong.Please ignore

starchan wrote: Clearly the equation gives $xabc < a$ I don't think that is true

But dividing by $abc$, gives $x=\frac{1}{bc} +\frac{1}{ca}+\frac{1}{ab} \le 3$, so $x \in \{1,2,3 \}$, noticing that the equations $a+b+c=xabc$ have solutions for all $x$ in the set $\{1,2,3\}$.

WLOG, suppose $a\leq b\leq c$. We then have $$xabc=a+b+c\leq 3c\implies xab\leq 3$$We can now proceed with casework on $(a,b)$ (there are not that many). Case 1: $(a,b)=(1,1)$ Our equation becomes $c+2=cx$. Since $c$ divides the LHS, $c=1,2$, which gives $x=3, 2$, respectively. Case 2: $(a,b)=(1,2)$ Our equation becomes $c+3=2cx$. Again, since $c$ divides the LHS, $c=1,3$. The former contradicts the maximality of $c$, but the latter gives $x=1$. Case 3: $(a,b)=(1,3)$ This gives $c+4=3cx$. Thus, $c=1,4$. The former is impossible due to the maximality of $c$, and the latter yields a contradiction mod 3. Having exhausted all cases, our solution sets for $(a,b,c,x)$ are $(1,1,2,2), (1,1,1,3), (1,2,3,1)$, with permutations on $a,b,c$.

Case 1: $a+b+c=abc$ Then, $a=3,b=2,c=1$, so $x=1$ works. Case 2: $a+b+c=2abc$ Then, $a=2,b=1,c=1$ works. Case 3: $a+b+c=3abc$ Then, $a=b=c=1$ works. Case 4: $a+b+c > 3abc$ Impossible. Note that $\frac{a+b+c}{abc} = \frac{1}{ab} + \frac{1}{ac} + \frac{1}{bc} \le 3$. Hence $x=1,2,3$ works & nothing else.