If $3a = 4b = 6c$ and $a + b + c = 27\sqrt{29}$, then $\sqrt{a^2 + b^2 + c^2}$ is

### Concept & Formula Use ratio proportions to express variables $a$, $b$, and $c$ in terms of a single common constant $k$, then substitute them into the given sum to solve for $k$. ### Step-by-Step Solution Let the given ratio equation be equal to a constant $k$: $$3a = 4b = 6c = k$$ Express $a$, $b$, and $c$ in terms of $k$: $$a = \frac{k}{3}, \quad b = \frac{k}{4}, \quad c = \frac{k}{6}$$ Substitute these values into the given sum equation $a + b + c = 27\sqrt{29}$: $$\frac{k}{3} + \frac{k}{4} + \frac{k}{6} = 27\sqrt{29}$$ Find the common denominator for the fractions, which is $12$: $$\frac{4k + 3k + 2k}{12} = 27\sqrt{29}$$ $$\frac{9k}{12} = 27\sqrt{29}$$ $$\frac{3k}{4} = 27\sqrt{29} \implies k = \frac{27\sqrt{29} \times 4}{3} = 36\sqrt{29}$$ Now, compute the values of $a$, $b$, and $c$: $$a = \frac{36\sqrt{29}}{3} = 12\sqrt{29}$$ $$b = \frac{36\sqrt{29}}{4} = 9\sqrt{29}$$ $$c = \frac{36\sqrt{29}}{6} = 6\sqrt{29}$$ Substitute these values into the required expression $\sqrt{a^2 + b^2 + c^2}$: $$\sqrt{(12\sqrt{29})^2 + (9\sqrt{29})^2 + (6\sqrt{29})^2}$$ $$\sqrt{29 \times (12^2 + 9^2 + 6^2)}$$ $$\sqrt{29 \times (144 + 81 + 36)} = \sqrt{29 \times 261}$$ Factorize $261$ to check for perfect squares: $261 = 29 \times 9$. $$\sqrt{29 \times 29 \times 9} = 29 \times 3 = 87$$ ### Exam Strategy & Shortcut Notice that $a : b : c = \frac{1}{3} : \frac{1}{4} : \frac{1}{6} = 4 : 3 : 2$. Let $a=4x, b=3x, c=2x$. Then $9x = 27\sqrt{29} \implies x = 3\sqrt{29}$. The value required is $\sqrt{(4x)^2 + (3x)^2 + (2x)^2} = \sqrt{29x^2} = x\sqrt{29}$. Substitute $x$: $(3\sqrt{29})\sqrt{29} = 3 \times 29 = 87$. ### Common Pitfall Avoid calculating out the massive numbers like $12^2 \times 29$. Keep the common radical term $\sqrt{29}$ factored outside the addition until the final step to keep calculations light. ### Final Answer **Therefore, the correct answer is 87.**