We need to find the number of real roots of $x|x-2|+3|x-3|+1=0.$

We will remove the absolute values by considering intervals based on the critical points $x=2$ and $x=3$.

1. Case-wise breakdown

The expression involves $|x-2|$ and $|x-3|$, so consider:

1. $x<2$
2. $2\le x<3$
3. $x\ge 3$

2. Case 1: $x<2$

Then $|x-2|=2-x,\qquad |x-3|=3-x.$ So the equation becomes $x(2-x)+3(3-x)+1=0.$ Expand: $2x-x^2+9-3x+1=0$ $-x^2-x+10=0$ $x^2+x-10=0.$ Its roots are $x=\frac{-1\pm\sqrt{1+40}}{2}=\frac{-1\pm\sqrt{41}}{2}.$ Now check which satisfy $x<2$:

• $\frac{-1+\sqrt{41}}{2}\approx 2.70>2$, not valid.
• $\frac{-1-\sqrt{41}}{2}\approx -3.70<2$, valid.

So this case gives one real root.

3. Case 2: $2\le x<3$

Then $|x-2|=x-2,\qquad |x-3|=3-x.$ So the equation becomes $x(x-2)+3(3-x)+1=0.$ Expand: $x^2-2x+9-3x+1=0$ $x^2-5x+10=0.$ Discriminant: $\Delta = (-5)^2-4(1)(10)=25-40=-15<0.$ Hence there are no real roots in this interval.

4. Case 3: $x\ge 3$

Then $|x-2|=x-2,\qquad |x-3|=x-3.$ So the equation becomes $x(x-2)+3(x-3)+1=0.$ Expand: $x^2-2x+3x-9+1=0$ $x^2+x-8=0.$ Its roots are $x=\frac{-1\pm\sqrt{1+32}}{2}=\frac{-1\pm\sqrt{33}}{2}.$ Now check which satisfy $x\ge 3$:

• $\frac{-1+\sqrt{33}}{2}\approx 2.37<3$, not valid.
• $\frac{-1-\sqrt{33}}{2}<0$, not valid.

So this case gives no real root.

5. Total number of real roots

Adding all valid roots from the three cases:

• Case 1: $1$
• Case 2: $0$
• Case 3: $0$

Therefore, the equation has $\boxed{1}$ real root.

6. Option matching

Given options:

• A: $4$
• B: $3$
• C: $2$
• D: $1$

So the correct option is $\boxed{\text{D}}.$

7. Comparison with stored answer

Stored correct answer: D

Our derived answer is also D, so they agree.