Assume, to the contrary, that the number of planes is equal to $1111$. Now, $20$ points in space can define at most $\binom{20}{3} = \frac{20 \cdot 19 \cdot 18}{3 \cdot 2 \cdot 1} = 1140$ planes, so $1140 - 1111 = 29$ triplets of points lie in the planes already determined by another triplet. If one of the planes contains $7$ or more points, then there are at least $\binom{7}{3} = 35$ triplets of points in this plane and the number of triplets is greater than the number of planes by at least $35 - 1 = 34$. Hence, the greatest possible number of planes is $1140 - 34 = 1105$. Obviously, this cannot happen if there are $1111$ planes, so each plane can contain at most $6$ of the points.

Let $a$ be the number of planes containing $4$ points, $b$ the number of planes containing $5$ points and $c$ the number of planes containing $6$ points. When counting triplets, we considered each plane containing $4$ points $\binom{4}{3} = 4$ times. That is $3$ times too many. Each plane containing $5$ points was counted $\binom{5}{3} = 10$ times (i.e. $9$ times too many) and each plane containing $6$ points was counted $\binom{6}{3} = 20$ times, which is $19$ times too many. The number of planes is thus equal to $1140 - 3a - 9b - 19c$. If this number were equal to $1111$, then we would have $3a + 9b + 19c = 29$. The numbers $a$, $b$ and $c$ are non-negative integers, so $c \ge 1$ is not possible, $c = 0$. We get $3a + 9b = 29$ and this is impossible since the left-hand side is divisible by $3$ and the right-hand side is not. We conclude that the number of planes cannot be equal to $1111$.