1. **State the problem:** Calculate the expression $$(3a+3b+3c)(27a+27b+27c) \div (9bc+9ca+9ab).$$ 2. **Rewrite the expression:** Factor out common terms: $$3(a+b+c) \times 27(a+b+c) \div 9(bc+ca+ab).$$ 3. **Simplify constants:** $$\frac{3 \times 27}{9} = \frac{81}{9} = 9,$$ so the expression becomes $$9 \times \frac{(a+b+c)^2}{bc+ca+ab}.$$ 4. **Recall the identity:** $$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).$$ 5. **Rewrite numerator:** $$9 \times \frac{a^2 + b^2 + c^2 + 2(ab + bc + ca)}{bc + ca + ab}.$$ 6. **Split the fraction:** $$9 \times \left( \frac{a^2 + b^2 + c^2}{bc + ca + ab} + 2 \right).$$ 7. **Final expression:** $$9 \left( 2 + \frac{a^2 + b^2 + c^2}{ab + bc + ca} \right).$$ This is the simplified form of the given expression.