Each center lies at height equal to its radius above its tangency point; write the second plane as $n \cdot x = c$ with $n$ a unit vector and express both tangencies

The conditions say the radius is an affine function of the tangency point in $\mathcal{P};$ it takes values $1, 2, 3$ at $A, B, C,$ so evaluate that function at $D$

A sphere of radius $\rho$ tangent to $\mathcal{P}$ at $P$ has center $P + \rho k,$ where $k$ is the upward unit normal of $\mathcal{P}.$ Write $\mathcal{T}$ as $\{x : n \cdot x = c\}$ with unit normal $n = (n_1, n_2, h)$ in coordinates where $\mathcal{P}$ is the $xy$-plane. Tangency with all spheres on the same side means $n \cdot (P + \rho k) - c = \rho$ for each sphere, that is $n_1 x_P + n_2 y_P - c = (1 - h)\rho.$ Here $h \ne 1,$ since otherwise the left side would be constant while the radii differ. So $\rho = g(P)$ for the affine function $g(x,y) = \frac{n_1 x + n_2 y - c}{1 - h}.$

Take $A = (0,0),$ $B = (6,0),$ $C = (0,4).$ The affine function with $g(A) = 1,$ $g(B) = 2,$ $g(C) = 3$ is $g(x,y) = 1 + \frac{x}{6} + \frac{y}{2}.$ The centroid is $G = \left(2, \frac{4}{3}\right),$ and line $BC$ is $2x + 3y = 12.$ Since $2 \cdot 2 + 3 \cdot \frac{4}{3} - 12 = -4,$ reflecting gives $$\begin{aligned} D &= G + \frac{8}{13}\,(2, 3) \\ &= \left(\frac{42}{13},\ \frac{124}{39}\right). \end{aligned}$$

Therefore $r = g(D)$ $= 1 + \frac{1}{6} \cdot \frac{42}{13}$ $+ \frac{1}{2} \cdot \frac{124}{39}$ $= 1 + \frac{21}{39} + \frac{62}{39}$ $= \frac{122}{39}.$ (Such a plane exists: the normal condition $(1-h)^2\left|\nabla g\right|^2 = 1 - h^2$ with $\left|\nabla g\right|^2 = \frac{5}{18}$ gives $h = -\frac{13}{23}.$) Since $\gcd(122, 39) = 1,$ the answer is $122 + 39 = 161.$