Assuming a digit is an element of $\{0,1,2,3,4,5,6,7,8,9,10\}$ we have three cases for $a,b,c$ to see:

1. $a=b=c=0$. All easy here, yields $1$ combination.
2. $b=c\ne 0$. $a=2b$, so $b<5$ giving us $4$ choices (digits $1$ to $4$). The position of $a$ uniquely determines the code, so multiply b $3$ to get $4\cdot 3 = 12$ combinations
3. $b\ne c$. We assume $a\ge b>c$ and chose $c$ first. Since $b+c < 10$ and $c<b$ $a \ge 2c$ so $c\le 4$. $$\begin{align*} c=4 & \Rightarrow b=5, a=9 & 1\\ c=3 & \Rightarrow b\in\{4,5,6\} & 3\\ c=2 & \Rightarrow b\in\{3,4,5,6,7\} & 5\\ c=1 & \Rightarrow b\in\{2,\ldots, 8\} & 7\\ c=0 & \Rightarrow b\in\{1,\ldots, 9\} & 9 (\text{only $2$ distinct digits here}) \end{align*}$$ totaling $14+9$ combinations, times $3! = 6$ for all but the $9$ we get $14\cdot 6 + 9 \cdot 3 = 111$

Summing up we have $1+12+111 = 124$ possibilities.

Assuming a digit is an element of $\{0,1,2,3,4,5,6,7,8,9,10\}$ we have three cases for $a,b,c$ to see:

1. $a=b=c=0$. All easy here, yields $1$ combination.
2. $b=c\ne 0$. $a=2b$, so $b<5$ giving us $4$ choices (digits $1$ to $4$). The position of $a$ uniquely determines the code, so multiply b $3$ to get $4\cdot 3 = 12$ combinations
3. $b\ne c$. We assume $a\ge b>c$ and chose $c$ first. Since $b+c < 10$ and $c<b$ $a \ge 2c$ so $c\le 4$. $$\begin{align*} c=4 & \Rightarrow b=5, a=9 & 1\\ c=3 & \Rightarrow b\in\{4,5,6\} & 3\\ c=2 & \Rightarrow b\in\{3,4,5,6,7\} & 5\\ c=1 & \Rightarrow b\in\{2,\ldots, 8\} & 7\\ c=0 & \Rightarrow b\in\{1,\ldots, 9\} & 9 (\text{only $2$ distinct digits here}) \end{align*}$$ totaling $16+9$ combinations, times $3! = 6$ for all but the $9$ we get $16\cdot 6 + 9 \cdot 3 = 123$

Summing up we have $1+12+123 = 136$ possibilities.