To determine if the equation

\[ \frac{a^2 + b^2 + c^2}{3} = \left(\frac{a + b + c}{3}\right)^2 \]

is generally true, let's analyze both sides of the equation.

First, let's simplify the right-hand side:

\[ \left(\frac{a + b + c}{3}\right)^2 \]

Expanding this, we get:

\[ \left(\frac{a + b + c}{3}\right)^2 = \frac{(a + b + c)^2}{9} \]

Now, let's expand \((a + b + c)^2\):

\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \]

So,

\[ \frac{(a + b + c)^2}{9} = \frac{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca}{9} \]

Now, let's compare this with the left-hand side of the original equation:

\[ \frac{a^2 + b^2 + c^2}{3} \]

For the equation to hold true, we need:

\[ \frac{a^2 + b^2 + c^2}{3} = \frac{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca}{9} \]

Multiplying both sides by 9 to clear the denominators, we get:

\[ 3(a^2 + b^2 + c^2) = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \]

Simplifying this, we have:

\[ 3a^2 + 3b^2 + 3c^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \]

Subtracting \(a^2 + b^2 + c^2\) from both sides, we get:

\[ 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca \]

Dividing both sides by 2, we obtain:

\[ a^2 + b^2 + c^2 = ab + bc + ca \]

This equation is not generally true for all \(a\), \(b\), and \(c\). Therefore, the original equation

\[ \frac{a^2 + b^2 + c^2}{3} = \left(\frac{a + b + c}{3}\right)^2 \]

is not generally true for all values of \(a\), \(b\), and \(c\).