Does this REMIND you of something else ?

Since when the polynomial $p(x)$ gives remainders $2$ and $1$ when it is divided by $x-1$ and $x-2$ respectively. We can assume $p(x)$ is of the form:

$p(x) = q(x)(x-1)(x-2) - x + 3$

$\Rightarrow r(x) = - (x-3)\quad \Rightarrow r(3) = - (3-3) = \boxed {0}$

Let $p(x)$ be the mysterious polynomial.

Now, knowing that $p(x)$ returns 1 & 2 when divided by $x-2$ and $x-1$ respectively, we can write $p(x)$ as :

$p(x) \equiv 2 \ \ mod (x-1) \quad and \quad p(x) \equiv 1 \ \ mod (x-2)$

Now, we assume that $deg (p) \geq 2$ (Indeed, as a side note, $p(x) = 3-x$ DOES solve the required properties up to now), in order to make the division by $(x-1)(x-2)$ sensible.

Hence, we may write $p(x)$ as : $p(x) = q(x)(x-1)(x-2) +r(x),$ where $q(x),r(x) \in \mathbb R [x]$ and $deg(r) \leq 1$ .

Hence, write $r(x) = ax+b$ for some $a,b \in \mathbb R$ Now, since $p(1) = 2 \ \ and \ \ p(2) = 1$ , we see that we require :

$r(2) = 1 \quad and \quad r(1) = 2$

Solving for $a$ and $b$ , we get : $\boxed{r(x) = 3-x}$ Hence : $\boxed{r(3) = 0}$

From the given, $P(x)$ can be expressed as:

$P(x)=(x-1)(x-2)Q(x)+R(x)$

$x-1=0 \rightarrow x=1, and x-2=0 \rightarrow x=2$

$P(1)=R(1)=2, and P(2)=R(2)=1$

Since the divisor $(x-1)(x-2)$ has a degree $2$ , $R(x)$ should be at most degree $1$ .

So, $R(x)=Ax+B$

From $R(1)=2, and R(2)=1$ ,

$R(1)=(1)A+B=A+B=2$

$R(2)=(2)A+B=2A+B=1$

Solving for A and B,

$A=-1, B=3$

Therefore, the remainder $R(x)$ is:

$R(x)=-x+3$

Solving for $R(3)$ ,

$R(3)=-(3)+3=\boxed0$