I will begin with a reformulation of your question which makes it not only more symmetric, by also (at least for me) more natural and interesting. I will pass from your variables $(W,H,Q)$ to new variables $(X,Y,Z)$ (which are not your original $X,Y,Z$) by putting $X=W, Y=-Q, Z=-H$. Then your condition (3) becomes just $X+Y+Z=0$. Therefore you are asking about a probability distribution $\pi$ on the plane $$\mathcal P=\{x+y+z=0\}\subset\mathbb R^3$$ such that all its one-dimensional marginals are symmetric, and you are interested in the possible values of $$ \begin{aligned} p_1=\pi\{x>0, y<0\} \;, \\ p_2=\pi\{y>0, z<0\} \;, \\ p_3=\pi\{z>0,x<0\} \;. \end{aligned} $$ Although it's not completely clear to me what is the continuity of random vectors you are referring to, I will presume that it implies that the probability $\pi$ of the line $\ell_x=\mathcal P \cap\{x=0\}$ and of the analogous lines $\ell_y,\ell_z\subset\mathcal P$ are all 0, so that replacing non-strict inequalities with strict ones doesn't change anything (the general case can also be treated in pretty much the same way). For a better visualization one can draw a picture of the plane $\mathcal P$ and the lines $\ell_x,\ell_y,\ell_z$ in the coordinates $x,y$ (so that $z=-x-y$).

Since all the domains from the definitions of the numbers $p_i$ are pairwise disjoint, and the complement of their union consists of the zero measure lines $\ell_x,\ell_y,\ell_z$, indeed $\sum p_1=1$. Moreover, the symmetry condition on the marginals means that all numbers $p_i$ satisfy the conditions $p_i\le \frac12$ (and, of course $p_i\ge 0$).

I claim that, conversely, any collection $(p_i)$ like this can be realized by a measure $\pi$ satisfying the above conditions. Indeed, take for $\pi$ the distribution with two atoms at the points $(1,-2,1)$ and $(-1,2,-1)$ and the weight $\frac12$. Then the corresponding $p$-vector $(p_1,p_2,p_3)$ is $$ v_3 = \left(\tfrac12, \tfrac12, 0\right) \;. $$ If you want to have an absolutely continuous measure $\pi$ with the same property, you can just smoothen the above distribution by taking the normalized sum of the uniform distributions on, say, side $\varepsilon$ squares centered at $(1,-2)$ and $(-1,2)$ in the coordinates $(x,y)$.

By symmetry, there are also measures $\pi$ with the $p$-vectors $v_2=\left(\tfrac12, 0, \tfrac12\right)$ and $v_1=\left(0, \tfrac12, \tfrac12\right)$. By taking convex combinations of the associated measures $\pi$ one can realize and $p$-vector from the convex hull of $v_1,v_2,v_3$, which is precisely the set determined by the above conditions on $p_i$ (for visualization one can draw a picture in the coordinates $p_1,p_2,1-p_1-p_2$).

I will begin with a reformulation of your question which makes it not only more symmetric, by also (at least for me) more natural and interesting. I will pass from your variables $(W,H,Q)$ to new variables $(X,Y,Z)$ (which are not your original $X,Y,Z$) by putting $X=W, Y=-Q, Z=-H$. Then your condition (3) becomes just $X+Y+Z=0$. Therefore you are asking about a probability distribution $\pi$ on the plane $$\mathcal P=\{x+y+z=0\}\subset\mathbb R^3$$ such that all its one-dimensional marginals are symmetric, and you are interested in the possible values of $$ \begin{aligned} p_1=\pi\{x>0, y<0\} \;, \\ p_2=\pi\{y>0, z<0\} \;, \\ p_3=\pi\{z>0,x<0\} \;. \end{aligned} $$ Although it's not completely clear to me what is the continuity of random vectors you are referring to, I will presume that it implies that the probability $\pi$ of the line $\ell_x=\mathcal P \cap\{x=0\}$ and of the analogous lines $\ell_y,\ell_z\subset\mathcal P$ are all 0, so that replacing non-strict inequalities with strict ones doesn't change anything (the general case can also be treated in pretty much the same way). For a better visualization one can draw a picture of the plane $\mathcal P$ and the lines $\ell_x,\ell_y,\ell_z$ in the coordinates $x,y$ (so that $z=-x-y$).

Since all the domains from the definitions of the numbers $p_i$ are pairwise disjoint, and the complement of their union consists of the zero measure lines $\ell_x,\ell_y,\ell_z$, indeed $\sum p_1=1$. Moreover, the symmetry condition on the marginals means that all numbers $p_i$ satisfy the conditions $p_i\le \frac12$ (and, of course $p_i\ge 0$).

I claim that, conversely, any collection $(p_i)$ like this can be realized by a measure $\pi$ satisfying the above conditions. Indeed, take for $\pi$ the distribution with two atoms at the points $(1,-2,1)$ and $(-1,2,-1)$ and the weight $\frac12$. Then the corresponding $p$-vector $(p_1,p_2,p_3)$ is $$ v_3 = \left(\tfrac12, \tfrac12, 0\right) \;. $$ If you want to have an absolutely continuous measure $\pi$ with the same property, you can just smoothen the above distribution by taking the normalized sum of the uniform distributions on, say, side $\varepsilon$ squares centered at $(1,-2)$ and $(-1,2)$ in the coordinates $(x,y)$.

By symmetry, there are also measures $\pi$ with the $p$-vectors $v_2=\left(\tfrac12, 0, \tfrac12\right)$ and $v_1=\left(0, \tfrac12, \tfrac12\right)$. By taking convex combinations of the associated measures $\pi$ one can realize any $p$-vector from the convex hull of $v_1,v_2,v_3$, which is precisely the set determined by the above conditions on $p_i$ (for visualization one can draw a picture in the coordinates $p_1,p_2,1-p_1-p_2$).