The points $\mathrm{A}(4,-2,1),\mathrm{B}(7,-4,7),\mathrm{C}(2,-5,10)$ and $\mathrm{D}(-1,-3,4)$ are the vertices of a

Here, the mid-point of AC is

$\left(\frac{4+2}{2},\frac{-2-5}{2},\frac{1+10}{2}\right)=\left(3,-\frac{7}{2},\frac{11}{2}\right)$

and that of BD is 

$\left(\frac{7-1}{2},\frac{-4-3}{2},\frac{7+4}{2}\right)=\left(3,-\frac{7}{2},\frac{11}{2}\right)$

So, the diagonals AC and BD bisect each other.

$\Rightarrow \mathrm{ABCD}$ is a parallelogram

$\text{ As }\left|\mathrm{AB}\right|=\sqrt{3^2+2^2+6^2}=7$ and $\left|\mathrm{AD}\right|=\sqrt{5^2+1^2+3^2}=\sqrt{35}\ne \left|\mathrm{AB}\right|$

Therefore, ABCD is not a rhombus and naturally, it cannot be a square.