1. Let the line cut the axes at $P(a,0),\quad Q(0,b)$ where $a>0, b>0$.

2. Equation of the line in intercept form $\frac{x}{a}+\frac{y}{b}=1$ Since the line passes through $(1,2)$, $\frac{1}{a}+\frac{2}{b}=1$

3. Area of triangle $OPQ$ Since $OP=a$ and $OQ=b$, $A=\frac{1}{2}ab$ We must minimize $ab$ subject to $\frac{1}{a}+\frac{2}{b}=1$

4. Express one variable in terms of the other From $\frac{1}{a}+\frac{2}{b}=1$ multiply by $ab$: $b+2a=ab$ Hence, $ab-b=2a$ $b(a-1)=2a$ $b=\frac{2a}{a-1}, \quad a>1$

5. Write area as a function of $a$ $A(a)=\frac{1}{2}a\cdot \frac{2a}{a-1}=\frac{a^2}{a-1}$

6. Differentiate and find critical point $A'(a)=\frac{2a(a-1)-a^2}{(a-1)^2}=\frac{a^2-2a}{(a-1)^2}=\frac{a(a-2)}{(a-1)^2}$ Setting $A'(a)=0$ gives $a=0 \text{ or } a=2$ Since $a>1$, the valid critical point is $a=2$

7. Find corresponding $b$ $b=\frac{2a}{a-1}=\frac{2\cdot 2}{2-1}=4$

8. Equation of the line The intercepts are $(2,0)$ and $(0,4)$, so slope is $m=\frac{4-0}{0-2}=\frac{4}{-2}=-2$

9. Check with options The slope is $-2$, which matches Option C.

Therefore, the required slope of $PQ$ is $\boxed{-2}$