The points (4, 7, 8), (2, 3, 4), (–1, –2, 1) and (1, 2, 5) are :

the vertices of a parallelogram

Let A ≡ (4, 7, 8), B ≡ (2, 3, 4), C ≡ (− 1, − 2, 1), D ≡ (1, 2, 5)

Direction cosines of AB ≡ $\left(\frac{2}{6}, \frac{4}{6}, \frac{4}{6}\right)=\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)$

Direction cosines of CD ≡ $\left(\frac{-2}{6}, \frac{-4}{6}, \frac{-4}{6}\right)$

$=\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)$

So, AB parallel to CD

Direction cosines of AD ≡ $\left(\frac{3}{\sqrt{43}}, \frac{5}{\sqrt{43}}, \frac{3}{\sqrt{43}}\right)$

Direction cosines of BC ≡ $\left(\frac{-3}{\sqrt{43}}, \frac{-5}{\sqrt{43}}, \frac{-3}{\sqrt{43}}\right)$

$=\left(\frac{3}{\sqrt{43}}, \frac{5}{\sqrt{43}}, \frac{3}{\sqrt{43}}\right)$

so, AD is parallel to BC.

Therefore ABCD is a parallelogram.