The expression provided is:
\[ (bc + ca + ab)^{10}. \]
The expansion's terms come from the multinomial expansion of \((bc + ca + ab)^{10}\). Let:
\[ x = bc, \quad y = ca, \quad z = ab. \]
The expansion of \((x + y + z)^{10}\) is:
\[ (x + y + z)^{10} = \sum_{i+j+k=10} \frac{10!}{i!j!k!} x^i y^j z^k. \]
Substitute back:
\[ x^i y^j z^k = (bc)^i (ca)^j (ab)^k = b^{i+k} c^{i+j} a^{j+k}. \]
We require:
\[ a^{10} \implies j + k = 10, \quad b^{7} \implies i + k = 7, \quad c^{3} \implies i + j = 3. \]
Solve these equations simultaneously:
From \(i + j = 3\), substitute \(j = 3 - i\) into \(j + k = 10\):
\[ 3 - i + k = 10 \implies k = 7 + i. \]
Substitute \(k = 7 + i\) into \(i + k = 7\):
\[ i + (7 + i) = 7 \implies 2i + 7 = 7 \implies i = 0. \]
Using \(i = 0\), calculate \(j\) and \(k\):
\[ j = 3 - i = 3, \quad k = 7 + i = 7. \]
The coefficient is given by:
\[ \frac{10!}{i!j!k!} = \frac{10!}{0! \cdot 3! \cdot 7!}. \]
Simplify:
\[ \frac{10!}{3! \cdot 7!} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120. \]
The coefficient of \(a^{10}b^{7}c^{3}\) is:
\[ \boxed{120}. \]