To prove that \(a = b = c\) given the equation \((a + b + c)^2 = 3(ab + bc + ac)\), let's start by expanding and simplifying both sides of the equation.

First, expand the left-hand side:
\[
(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca
\]

Next, consider the right-hand side:
\[
3(ab + bc + ca)
\]

Equate the two expressions:
\[
a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 3(ab + bc + ca)
\]

Now, simplify the equation by subtracting \(3(ab + bc + ca)\) from both sides:
\[
a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 3(ab + bc + ca) = 0
\]

Combine like terms:
\[
a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca = 0
\]

Simplify further:
\[
a^2 + b^2 + c^2 - ab - bc - ca = 0
\]

Rearrange the equation:
\[
a^2 + b^2 + c^2 = ab + bc + ca
\]

To prove that \(a = b = c\), let's assume \(a\), \(b\), and \(c\) are not all equal and see if we can derive a contradiction. Without loss of generality, assume \(a \geq b \geq c\).

Consider the case where \(a = b\). Then the equation becomes:
\[
a^2 + a^2 + c^2 = a^2 + ac + ac
\]
\[
2a^2 + c^2 = a^2 + 2ac
\]
\[
a^2 + c^2 = 2ac
\]
\[
a^2 - 2ac + c^2 = 0
\]
\[
(a - c)^2 = 0
\]
\[
a = c
\]

Since \(a = b\) and \(a = c\), we have \(a = b = c\).

Now consider the case where \(a \neq b\) and \(b \neq c\). We can rewrite the equation as:
\[
a^2 + b^2 + c^2 - ab - bc - ca = 0
\]

This equation is symmetric in \(a\), \(b\), and \(c\). If \(a\), \(b\), and \(c\) are distinct, we can use the method of symmetric sums. However, the simplest way to see this is to note that the equation \(a^2 + b^2 + c^2 - ab - bc - ca = 0\) implies that the sum of the squares of the differences between the variables must be zero. Specifically:
\[
\frac{1}{2} \left( (a - b)^2 + (b - c)^2 + (c - a)^2 \right) = 0
\]

Since the sum of squares is zero, each individual square must be zero:
\[
(a - b)^2 = 0, \quad (b - c)^2 = 0, \quad (c - a)^2 = 0
\]

Thus:
\[
a = b = c
\]

Therefore, we have shown that \(a = b = c\) is the only solution that satisfies the given equation.