For any real number y, y² 0.

For any real number $y$, $y^2 \ge 0$.
Therefore, for any real value of $x$:

The given equation is $(x-1)^2 + (x-2)^2 + (x-3)^2 = 0$.
Since each term is non-negative, their sum can only be zero if and only if each individual term is equal to zero.
This implies:

For the original equation to be true, $x$ must simultaneously satisfy $x=1$, $x=2$, and $x=3$. This is impossible, as $x$ cannot be three different values at the same time.

Since there is no real value of $x$ that can satisfy all conditions simultaneously, the equation has no real roots.

The number of real roots is 0.

The final answer is $\text{0}$.