If 3a = 4b = 6c and a + b + c = 27sqrt(29) find sqrt(a ^ 2 + b ^ 2 + c ^ 2)

We are given the system of equations:

\[
3a = 4b = 6c
\]

and

\[
a + b + c = 27\sqrt{29}
\]

### Step 1: Express \(a\), \(b\), and \(c\) in terms of one variable

From the equation \(3a = 4b = 6c\), we can introduce a variable \(k\) such that:

\[
3a = 4b = 6c = k
\]

This gives us the following expressions:

\[
a = \frac{k}{3}, \quad b = \frac{k}{4}, \quad c = \frac{k}{6}
\]

### Step 2: Use the equation \(a + b + c = 27\sqrt{29}\)

Substitute \(a = \frac{k}{3}\), \(b = \frac{k}{4}\), and \(c = \frac{k}{6}\) into the equation \(a + b + c = 27\sqrt{29}\):

\[
\frac{k}{3} + \frac{k}{4} + \frac{k}{6} = 27\sqrt{29}
\]

To add these fractions, find the least common denominator (LCD) of 3, 4, and 6, which is 12:

\[
\frac{k}{3} = \frac{4k}{12}, \quad \frac{k}{4} = \frac{3k}{12}, \quad \frac{k}{6} = \frac{2k}{12}
\]

Now add the fractions:

\[
\frac{4k}{12} + \frac{3k}{12} + \frac{2k}{12} = \frac{9k}{12} = \frac{3k}{4}
\]

So, we have:

\[
\frac{3k}{4} = 27\sqrt{29}
\]

### Step 3: Solve for \(k\)

Multiply both sides by 4:

\[
3k = 108\sqrt{29}
\]

Now divide by 3:

\[
k = 36\sqrt{29}
\]

### Step 4: Find \(\sqrt{a^2 + b^2 + c^2}\)

Now, we need to compute \(\sqrt{a^2 + b^2 + c^2}\). Using the expressions for \(a\), \(b\), and \(c\):

\[
a = \frac{k}{3} = \frac{36\sqrt{29}}{3} = 12\sqrt{29}
\]
\[
b = \frac{k}{4} = \frac{36\sqrt{29}}{4} = 9\sqrt{29}
\]
\[
c = \frac{k}{6} = \frac{36\sqrt{29}}{6} = 6\sqrt{29}
\]

Now compute \(a^2 + b^2 + c^2\):

\[
a^2 = (12\sqrt{29})^2 = 144 \times 29 = 4176
\]
\[
b^2 = (9\sqrt{29})^2 = 81 \times 29 = 2349
\]
\[
c^2 = (6\sqrt{29})^2 = 36 \times 29 = 1044
\]

Now add these:

\[
a^2 + b^2 + c^2 = 4176 + 2349 + 1044 = 7569
\]

### Step 5: Take the square root

\[
\sqrt{a^2 + b^2 + c^2} = \sqrt{7569} = 87
\]

Thus, the value of \(\sqrt{a^2 + b^2 + c^2}\) is \(87\).

---

Would you like any further details or clarifications on the steps?

### Related questions:
1. How would the solution change if the values of \(a + b + c\) were different?
2. How can this method be applied to solve other systems of proportional equations?
3. What is the significance of introducing the variable \(k\) in solving this type of equation?
4. Can this approach be generalized for \(ma = nb = pc\) type of systems?
5. How do different numerical values of \(a\), \(b\), and \(c\) affect the value of \(\sqrt{a^2 + b^2 + c^2}\)?

### Tip:
When dealing with proportional equations, using a common variable like \(k\) can simplify the process of solving for multiple unknowns at once.

If 3a = 4b = 6c and a + b + c = 27√29, find √(a^2 + b^2 + c^2)

Learn how to solve the proportional system of equations 3a = 4b = 6c and determine the value of √(a^2 + b^2 + c^2) when a + b + c = 27√29. This problem covers algebraic concepts such as proportional relationships and radical expressions, suitable for students in Grades 10-12.

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