Modular Inverses

You can compute that a = 27, so the remainder is 5.

Note that $11 \times 1 + 1= 12 = 3 \times 4$, so 3 and 4 are multiplicative inverses of each other mod 11.

Similarly, we have $11 \times 4 + 1 = 45 = 5 \times 9$ so 5 and 9 are multiplicative inverses of each other mod 11.

Also, $11 \times 5 + 1 = 56 = 7 \times 8$ so 7 and 8 are multiplicative inverses of each other mod 11.

Therefore, $a \equiv (3^{-1} + 5^{-1} + 7^{-1})^{-1} \equiv (4 + 9 + 8)^{-1} \equiv 21^{-1} \pmod{11} \equiv -1^{-1} \equiv 10 \pmod{11}$.