jmc

Since a parabola can be tangent to a given line in at most one point, the parabola must be tangent to all three lines $y=-11x-37,$ $y=x-1,$ and $y=9x+3.$ Thus, if $a$ is the leading coefficient of $p(x),$ then $$\begin{array}{rl}p(x)-(-11x-37)& =a(x-x_1{)}^2,\\ p(x)-(x-1)& =a(x-x_2{)}^2,\\ p(x)-(9x+3)& =a(x-x_3{)}^2.\end{array}$$Subtracting the first two equations, we get $$\begin{array}{rl}12x+36& =a(x-x_1{)}^2-a(x-x_2{)}^2\\ & =a(x-x_1+x-x_2)(x_2-x_1)\\ & =2a(x_2-x_1)x+a(x_1^2-x_2^2).\end{array}$$Matching coefficients, we get $$\begin{array}{rl}2a(x_2-x_1)& =12,\\ a(x_1^2-x_2^2)& =36.\end{array}$$Dividing these equations, we get $-\frac{1}{2}(x_1+x_2)=3,$ so $x_1+x_2=-6.$

Subtracting other pairs of equations gives us $x_1+x_3=-4$ and $x_2+x_3=-1.$ Then $2x_1+2x_2+2x_3=-11,$ so $$x_1+x_2+x_3=\boxed{-\frac{11}{2}}.$$