Let me note that you are miscounting the possibilties, under any of two interpretations I can give to your description.
As you note, you have $2^3$ ways of filling in the 1st, 4th, and 5th position of the number. And you have $2^2$ ways of filling in the 3rd and 10th. So it comes down to figuring out the number of ways to fill out the remaining five positions. To count those, I will ignore the 1st, 3rd, 4th, 5th, and 10th position, and think simply that we are trying to fill out five positions (which will correspond to the 2nd, 6th, 7th, 8th, and 9th positions of the phone number).
If exactly two of these five positions must be equal to $0$, then we can first select which two positions will get the zeros: this can be done in $\binom{5}{2}=10$ ways. Then we select the digits to go in the three remaining positions, in order: there are $9$ possible digits for the first open position, $9$ for the second open position, and $9$ for the third open position. This gives $10\times 9^3$, and that's it. We don't need to "shuffle them". In fact, we shouldn't! If we multiply by $5!$, then we are overcounting: consider the situation in which we select the first two positions for the $0$, and we take 1, 2, and 3 for the remaining three open positions, in that order. Multiplying by 5! will give us all possible "reshuffles" of 00123; e.g., 00312. But we already counted that one: we counted it when we said that 3 could be in the first open position, 1 in the second, and 2 in the third. Worse! Multiplying by 5! includes the shuffle in which we exchange the first and second positions and do nothing else: but doing that to 00123 just gives 00123 back! So we are overcounting. We don't need that 5!.
Under this interpretation ("exactly two zeros"), the total number (ignoring the restriction on the sum being equal to 42$) would be $2^3\times 2^2\times\binom{5}{2}\times 9^3 = 2^5\times 10\times 9^3.
Now consider the case in which there are at least two zeros, but could be more. In this case, after selecting the two positions for the two zeros, it would seem that we now have 10 choices for each of the remaining three open slots, giving \binom{10}{2}\times 10^3 = 10^4$ possibilities. Unfortunately, this overcounts as well, when the number has more than two zeros. For example, 00012 is counted three times: once when we designate the first and second positions as having $0s, and happen to select the third as having a zero; again when we designate first and third positions; and yet again when we designate second and third positions. So you would have to be more carefl; here as well.
One possibility is to use inclusion-exclusion; another is to be more careful, and count those with exactly two zeros first; then those with exactly three zeros (\binom{5}{3}\times 9^2$); then those with exactly four zeros ($\binom{5}{4}\times 9$), and finally those with exactly five zeros ($\binom{5}{5}=1$). In any case, we don't need that $5!$ "shuffling factor" here either.