The points (5, - 4,2),(4,- 3, 1),(7, - 6, 4) and (8, - 7 , 5) are the vertices of

Let , A(5, -4,2), B(4, - 3, 1), C(7, -6, 4) and D(8, - 7 ,5). 

Then,

$\begin{array}{r}\mathrm{AB}=\sqrt{(4-5{)}^2+(-3+4{)}^2+(1-2{)}^2}\\ =\sqrt{1+1+1}=\sqrt{3}\\ \mathrm{BC}=\sqrt{(7-4{)}^2+(-6+3{)}^2+(4-1{)}^2}\\ =\sqrt{9+9+9}=3\sqrt{3}\\ \mathrm{CD}=\sqrt{(8-7{)}^2+(-7+6{)}^2+(5-4{)}^2}\\ =\sqrt{1+1+1}=\sqrt{3}\\ \mathrm{AD}=\sqrt{(8-5{)}^2+(-7+4{)}^2+(5-2{)}^2}\\ =\sqrt{9+9+9}=3\sqrt{3}\\ \mathrm{AC}=\sqrt{(7-5{)}^2+(-6+4{)}^2+(4-2{)}^2}\\ =\sqrt{4+4+4}=2\sqrt{3}\\ \mathrm{BD}=\sqrt{(8-4{)}^2+(-7+3{)}^2+(5-1{)}^2}\\ =\sqrt{16+16+16}=4\sqrt{3}\end{array}$

Here, we see that opposite sides .AB = CD,BC = AD are equal and diagonals $\text{AC≠BD}$
Hence, given points are the vertices of a parallelogram.