What you did is not correct. Note that the sums of the probabilities you considered is not $1$.
When we are dealing with conditional expectation, we deal with a new probability measure on the space. In this case, it's the probability $$P(\cdot \mid X \in \{1,2\}).$$
Therefore, the conditional expectation is $$E(X \mid X\in\{1,2\} = 1 \times P(X=1 \mid X\in \{1,2\})+2\times P(X=2 \mid X\in \{1,2\}) + 3 \times P(X=3 \mid X\in \{1,2\})=1\times \frac{1/3}{5/6}+2 \times \frac{1/2}{5/6}+3 \times 0 = \frac{8}{5}.$$$$E(X \mid X\in\{1,2\}) = 1 \times P(X=1 \mid X\in \{1,2\})+2\times P(X=2 \mid X\in \{1,2\}) + 3 \times P(X=3 \mid X\in \{1,2\})=1\times \frac{1/3}{5/6}+2 \times \frac{1/2}{5/6}+3 \times 0 = \frac{8}{5}.$$
I hope this helps you.