Equation of given curve is
$y=x^4-6 x^3+13 x^2-10 x+5$
So, $\quad \frac{d y}{d x}=4 x^3-18 x^2+26 x-10$
As the slope of tangent at point $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right)$ on curve is given as 2 , so put
$\frac{d y}{d x}=2$
$\Rightarrow \quad 4 x^3-18 x^2+26 x-10=2$
$\Rightarrow \quad 2 x^3-9 x^2+13 x-6=0$
$\Rightarrow \quad(x-1)\left(2 x^2-7 x+6\right)=0$
$\Rightarrow \quad(x-1)\left(2 x^2-4 x-3 x+6\right)=0$
$\Rightarrow \quad(x-1)[2 x(x-2)-3(x-2)]=0$
$\Rightarrow \quad(x-1)(x-2)(2 x-3)=0$
$\therefore \quad x=1, \frac{3}{2}, 2 \therefore\left(x_1, x_2\right)=(1,2)$,
$\because \quad x_1, x_2 \in \mathbf{N} \Rightarrow y_1=3$ and $y_2=5$
Therefore, $x_1 x_2+y_1 y_2=(1 \times 2)+(3 \times 5)$
$=2+15=17$