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

Let a line$AC$  divides at $B$ in ratio $m:n$

$A(2,3,4),B\left(-1,-2,1\right) \& C\left(5,8,7\right)$

So $(-1,2,1)=\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n},\frac{m{z}_2+n{z}_1}{m+n}$

$\Rightarrow \left(-1,2,1\right)=\frac{5m+2n}{m+n},\frac{8m+n}{m+n},\frac{7m+4n}{m+n}$

$\Rightarrow -1=\frac{5m+2n}{m+n},-2=\frac{8m+n}{m+n},1=\frac{7m+4n}{m+n}$

$\Rightarrow \frac{m}{n}=\frac{-1}{2},\frac{m}{n}=\frac{-1}{2},\frac{m}{n}=\frac{-1}{2}$

Ratio are same. It means points are collinear.