Solving the Algebraic Equation

The given equation is \(a^2 + b^2 + c^2 + 189 = 6(a + 4b + 2c)\). Our goal is to find the value of \(\sqrt{ab + bc + ca}\).

Rearranging and Completing the Square

First, rearrange the equation to group terms involving \(a\), \(b\), and \(c\):

\(a^2 + b^2 + c^2 + 189 = 6a + 24b + 12c\)

\(a^2 - 6a + b^2 - 24b + c^2 - 12c + 189 = 0\)

Now, complete the square for each variable:

• For \(a\): \((a^2 - 6a + 9)\)
• For \(b\): \((b^2 - 24b + 144)\)
• For \(c\): \((c^2 - 12c + 36)\)

The sum of the constants required is \(9 + 144 + 36 = 189\). Substitute these back into the equation:

\((a^2 - 6a + 9) + (b^2 - 24b + 144) + (c^2 - 12c + 36) + 189 - (9 + 144 + 36) = 0\)

This simplifies to:

\((a - 3)^2 + (b - 12)^2 + (c - 6)^2 = 0\)

Determining Variable Values

Since the sum of squares of real numbers is zero if and only if each term is zero, we have:

• \(a - 3 = 0 \implies a = 3\)
• \(b - 12 = 0 \implies b = 12\)
• \(c - 6 = 0 \implies c = 6\)

Calculating \(ab + bc + ca\)

Substitute the values of \(a\), \(b\), and \(c\) to find \(ab + bc + ca\):

\(ab = (3)(12) = 36\)

\(bc = (12)(6) = 72\)

\(ca = (6)(3) = 18\)

\(ab + bc + ca = 36 + 72 + 18 = 126\)

Finding the Square Root

Finally, calculate \(\sqrt{ab + bc + ca}\):

\(\sqrt{126}\)

Simplify the radical:

\(\sqrt{126} = \sqrt{9 \times 14} = \sqrt{3^2 \times 14} = 3\sqrt{14}\)

Thus, \(\sqrt{ab + bc + ca} = 3\sqrt{14}\).