Foundations · Superchallenge problem

The expression $(x - 2y + 3z) - (2x - 3y + 4z) + (3x - 4y + 5z) - \cdots + (99x - 100y + 101z)$ has $99$ groups, with signs alternating $+, -, +, \ldots$ and the $k$th group equal to $kx - (k+1)y + (k+2)z$. Evaluate it when $x = 3$, $y = 2$, $z = 1$.