Solution:

Suppose we start with $a$ $0$s, $b$ $1$s and $c$ $2$s. Each move changes the parity of all of $a$, $b$, $c$. Each move reduces the length of the sequence by $1$, so there must be $a + b + c - 1$ moves in all. Hence the total number of $0$s, the total number of $1$s and the total number of $2$s all have their parity changed $a + b + c - 1$ times. So if we end up with just one $0$, then $a$ must have the opposite parity to $b$ and $c$. In that case, we cannot end up with just one $1$, or just one $2$. Similarly in the other cases.