Answer: 71

Solution:

Let $a$ be the common root of $P+Q$ and $P+R,$ let $b$ be that of $P+Q$ and $Q+R,$ and let $c$ be that of $P+R$ and $Q+R.$ Each sum is quadratic with leading coefficient $2,$ so $P+Q = 2(x-a)(x-b),$ $P+R = 2(x-a)(x-c),$ and $Q+R = 2(x-b)(x-c).$ Write $Q = x^2 + qx + 2$ and $R = x^2 + rx + s.$ By Vieta's formulas, the root sums are $a + b = \frac{3-q}{2},$ $a + c = \frac{3-r}{2},$ and $b + c = -\frac{q+r}{2},$ so $$\begin{aligned} 2a &= (a+b) + (a+c) - (b+c) \\ &= \frac{3-q}{2} + \frac{3-r}{2} \\ &\quad {}+ \frac{q+r}{2} = 3, \end{aligned}$$ giving $a = \frac{3}{2}.$

The constant term of $P + Q$ is $-7 + 2 = -5,$ so $2ab = -5$ and $b = -\frac{5}{3}.$ Also $2ac = P(0) + R(0) = s - 7$ and $2bc = Q(0) + R(0) = s + 2.$ Subtracting, $2c(b - a) = 9,$ so $c = \frac{9}{2\left(-\frac{5}{3} - \frac{3}{2}\right)} = -\frac{27}{19}.$

Then $s = 2bc - 2$ $= 2 \cdot \left(-\frac{5}{3}\right)\left(-\frac{27}{19}\right) - 2$ $= \frac{90}{19} - \frac{38}{19} = \frac{52}{19},$ so $R(0) = \frac{52}{19}$ and $m + n = 52 + 19 = 71.$