If the system x+y+z=1, x+2y+4z=k, x+4y+10z=k^2 has infinitely many solutions, then the sum of all possible values of k is

The coefficient determinant is $1(20-16)-1(10-4)+1(4-2)=0$, so solutions are either none or infinitely many. Eliminating $x$: subtracting the first equation from the second and third gives $y+3z=k-1$ and $3y+9z=k^2-1$. The second is three times the first on the left, so consistency needs $k^2-1=3(k-1)$, i.e. $k^2-3k+2=0$, so $k=1$ or $k=2$. For both values the system reduces to two independent equations in three unknowns. The sum is $3$; the sum of squares, $5$, and the product, $2$, are the distractors.