a) Step 1: Analyze the properties of squared terms. For any real number $y$, $y^2 \ge 0$. Therefore, for any real value of $x$: $$(x-1)^2 \ge 0$$ $$(x-2)^2 \ge 0$$ $$(x-3)^2 \ge 0$$ Step 2: Evaluate the condition for the sum to be zero. 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: $$(x-1)^2 = 0 \quad \Rightarrow \quad x-1 = 0 \quad \Rightarrow \quad x = 1$$ $$(x-2)^2 = 0 \quad \Rightarrow \quad x-2 = 0 \quad \Rightarrow \quad x = 2$$ $$(x-3)^2 = 0 \quad \Rightarrow \quad x-3 = 0 \quad \Rightarrow \quad x = 3$$ Step 3: Check for a common solution. 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. Step 4: Determine the number of real roots. 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 $\boxed{\text{0}}$.