Three fair dice are thrown together, so there are 216 equally likely ordered outcomes…

Step 1. Total 10. The ways for the totals $3$ to $18$ are $1, 3, 6, 10,$ $15, 21, 25, 27, 27,$ $25, 21, \dots$. The total $10$ is the eighth entry, so it has $27$ ways. To check it, count with $x_1 + x_2 + x_3 = 10$: $^9C_2 - 3 \times {}^3C_2 = 36 - 9 = 27$.

Step 2. Total 13. By symmetry the total $13$ has the same ways as the total $21 - 13 = 8$, which is $21$. Check: $^{12}C_2 - 3 \times {}^6C_2 = 66 - 45 = 21$.

Step 3. At most 6. The totals $3$, $4$, $5$ and $6$ have $1$, $3$, $6$ and $10$ ways, so $20$ outcomes. The probability is $\dfrac{20}{216} \approx 0.0926$.