For which value of $k$ does the system $x + y + z = 1$, $x + 2y + 3z = 2$, $x + 3y + kz = 3$ have infinitely many solutions?

Explanation

1. 01

   A unique solution fails when $|A| = 0$: $|A| = 1(2k - 9) - 1(k - 3) + 1(3 - 2) = k - 5$, so $k = 5$.

2. 02

   With $k = 5$, the third equation is twice the second minus the first, and so is its right side ($2 \times 2 - 1 = 3$): the system is consistent, with infinitely many solutions.