Note that the expression evaluates to $3a^2+b^2+c^2=2023$. It turns out that this equation has only one solution $(a,b,c)=(23,20,6)$ which satisfies $a>b>c$ and the Triangle Inequality and hence all possible perimters of $ABC$ is $23+20+6=49$.

What a beautiful solution! Didn't get this in the exam. Hats off!

What exam is this problem from ?

we have $T_{1}=(a+b+c)^2-(a-b+c)^2+(a-b-c)^2-(a+b-c)^2 =(a+b+c+a-b+c)(a+b+c-a+b-c)+(a-b-c+a+b-c)(a-b-c-a-b+c)=(2a+2c)(2b)+(2a-2c)(-2b)=4bc$ $T_{2}=(a+b+c)^{4}-(a-b+c)^{4}+(a-b-c)^{4}-(a+b-c)^{4}=\left((a+b+c)^{2}+(a-b+c)^{2}\right)\left((a+b+c)^2-(a-b+c)^{2}\right)+\left((a-b-c)^2+(a+b-c)^2\right) \left((a-b-c)^{2}-(a+b-c)^{2}\right)=2b(2a+2c)(2a^2+2b^2+2c^2+4ac)+(2a+2c)(-2b)(2a^2+2b^2+2c^2-4ac)=2b\left((2a^2+2b^2+2c^2)(2c)+4ac(2a)\right)$ so we get $T_{2}=4bc\left(6a^2+2b^2+2c^2\right)$ , so we get $\frac{T_{2}}{2T_{1}}=3a^2+b^2+c^2$ so we need to solve $3a^2+b^2+c^2= 2023 $ for positive integers forming a triangle and $a>b>c$ , clearly $a^2 \equiv 1 \pmod 4, b^2, c^2 \equiv 0 \pmod 4$ and $b^2+c^2=2023-3a^2 \implies a \leqslant 25$ which gives only possible triplet as $(a,b,c)=(23,20,6)$ , so perimeter of this triangle is $\boxed{49}$. $\blacksquare$