1. **State the problem:** Calculate the expression $$\frac{(3a+3b+3c)(27a+27b+27c)}{9bc+9ca+9ab}$$. 2. **Rewrite the expression:** Factor out constants from each term: $$3(a+b+c) \times 27(a+b+c) \div 9(bc+ca+ab)$$. 3. **Simplify constants:** $$\frac{3 \times 27 (a+b+c)^2}{9(bc+ca+ab)} = \frac{81 (a+b+c)^2}{9(bc+ca+ab)}$$. 4. **Divide constants:** $$\frac{81}{9} = 9$$, so the expression becomes: $$9 \times \frac{(a+b+c)^2}{bc+ca+ab}$$. 5. **Recall the expansion:** $$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$. 6. **Rewrite the expression:** $$9 \times \frac{a^2 + b^2 + c^2 + 2(ab + bc + ca)}{bc + ca + ab}$$. 7. **Split the fraction:** $$9 \times \left( \frac{a^2 + b^2 + c^2}{ab + bc + ca} + 2 \right)$$. **Final answer:** $$9 \left( 2 + \frac{a^2 + b^2 + c^2}{ab + bc + ca} \right)$$.