Prove $\{abc : a+b=c\}$ is not context-free using pumping lemma
Suppose that $L$ were context-free. Then $L \cap 10^*20^*30^*$ would also be context-free. What is this language? It is easy to see that each of $a,b,c$ needs to contain a non-zero digit. Therefore $a = 10^i$, $b = 0^j20^k$, and $c=0^\ell30^m$. Clearly $i=k=m$, and so
$$L \cap 10^*20^*30^* = \{10^{i+j}20^{i+\ell}30^i : i,j,\ell \geq 0\} = \{10^i20^j30^k : i,j \geq k\}.$$

To make our life easier, let us consider instances with more structure: $L \cap 30^*130^*160^*2$. Consider any word $30^i130^j160^k2$. The only way in which $a+b$ ends with $2$ is if $a,b$ end with $1$, and so $a=30^i1$, $b=30^j1$, $c=60^j2$. This shows that
$$
L' \triangleq L \cap 30^*1 30^*1 60^*2 = \{ 30^i130^i160^i2 : i \geq 0\}.
$$
If $L$ were context-free, so would $L'$ be. Now define a homomorphism $h$ which maps $1,2,3,6$ to themselves and $a,b,c$ to $0$. Clearly
$$
L'' \triangleq h^{-1}(L') \cap 3a^*13b^*16c^*2 = \{3a^i13b^i16c^i2 : i \geq 0\}.
$$
If $L'$ were context-free so would $L''$ be. Finally, define a homomorphism $k$ which erases $1,2,3,6$ and fixes $a,b,c$. Then the following language would also be context-free:
$$
k(L'') = \{ a^i b^i c^i : i \geq 0 \}.
$$
However, this language is well-known to be non-context-free.



If for some reason you want to prove that $L$ is not context-free using the pumping lemma (the only reason I can see is that you were given this as an exercise), then you could consider instances of the form $30^n130^n160^n2$, which might be easier to handle than your original suggestion.