Let $A\equiv \left(2,-1,3\right)$, $B\equiv \left(4,-2,1\right)$, $C\equiv \left(4,5,-7\right)$ & $D\equiv \left(2,6,-5\right)$

Let mid-point of $AC$ & $BD$ are $M$ & $N$

$M=\left(\frac{4+2}{2},\frac{5-1}{2},\frac{-7+3}{2}\right)$, $N=\left(\frac{4+2}{2},\frac{-2+6}{2},\frac{1-5}{2}\right)$

$\Rightarrow M=N=\left(3,2,-2\right)$

So, at least quadrilateral is a parallelogram.

$\overrightarrow{AB}=2\hat{i}-\hat{j}-2\hat{k}$

$\overrightarrow{BC}=7\hat{j}-8\hat{k}$

$\overrightarrow{AB}.\overrightarrow{BC}=-7+16\ne 0$

$\Rightarrow AB$ is not perpendicular to $BC$.

$AB=\sqrt{{\left(4-2\right)}^2+{\left(-2+1\right)}^2+{\left(1-3\right)}^2}$

$\Rightarrow AB=3$

$BC=\sqrt{{\left(4-4\right)}^2+{\left(5+2\right)}^2+{\left(-7-1\right)}^2}$

$\Rightarrow BC=\sqrt{113}$

$AB\ne BC$

So, $ABCD$ is a parallelogram.