No one of the three numbers is even. Next $a,b,c$ have same residue modulo $4$ as $a^2bc,ab^2c,abc^2$ are all $3$ mod $4$ and then we get a contradiction.

A little different than the above one. If any of $a,b,c$ is even, one of these is $2 \pmod{4}$ whiich is impossible. So, all are odd. Therefore, $a^2,b^2,c^2$ are all $1\pmod{4}$ and also, the given square numbers are odd. So, $bc+2,ca+2,ab+2$ all are $2 \pmod{4}$. Therefore, $ab,bc,ca$ are all $3 \pmod{4}$. So, $(abc)^2=(ab)(bc)(ca) \equiv 3^3 \equiv 3 \pmod{4}$, contradiction.

None of $a$ , $b$ as well as $c$ is even.If there is one,then there is a number congruent to $2 (mod4)$ among the 3 given the numbers.Thus all of them cannot be squares.Now as they are all odd numbers there are only 2 congruent modulo 4 to be more specific $1,-1(mod4).$But there are $3$ numbers.So by PHP two numbers have the same congruent (mod4).Let $a$ and $b$ be those 2 numbers(WLOG).Then by mere inspection,$abc^2+2$ is congruent to $3(mod4)$.Contra!Done.

first we take abc=k then ka + 2,kb +2,kc+ 2 are all squares..(according to the problem) and now if all of a,b,c are even then k=even and ka = 2mod4 ,kb = 2mod4,kc=2mod4.....again if only two of them are even then there exists two of the above mentioned numbers which are congruent to 2modulo4 hence proving to non squares.................again if only one of a,b,c are even then also one of them turns out to be a nonsquare...........so all of them are odd. now as all of them are odd i.e they are either congruent to 1,-1 modulo 4............now taking any of them without loss of generality now it must be 1,-1mod4 if it is 1mod4 then either b,c=-1mod4 or only one of b,c=1mod4 or both 1mod 4 so taking both of the nos with same modulo4 and checking gives us that the nos ka,kb,kc one of them is 3mod4.....................adios amigos