Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6),$ and the product of the radii is $68.$ The x-axis and the line $y=mx$, where $m>0,$ are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt{b}/c,$ where $a,$ $b,$ and $c$ are positive integers, $b$ is not divisible by the square of any prime, and $a$ and $c$ are relatively prime. Find $a+b+c.$
1. **Identify the properties of the circles:**
   - The circles \(\mathcal{C}_1\) and \(\mathcal{C}_2\) intersect at \((9,6)\).
   - The product of the radii of the circles is \(68\).
   - Both circles are tangent to the x-axis and the line \(y = mx\).

2. **Consider a general circle tangent to the x-axis and the line \(y = mx\):**
   - Let the center of the circle be \((a, r)\), where \(r\) is the radius.
   - The distance from the center to the x-axis is \(r\), so the circle is tangent to the x-axis.
   - The distance from the center to the line \(y = mx\) must also be \(r\).

3. **Calculate the distance from the center \((a, r)\) to the line \(y = mx\):**
   - The distance from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by:
     \[
     \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
     \]
   - For the line \(y = mx\), we can rewrite it as \(mx - y = 0\), so \(A = m\), \(B = -1\), and \(C = 0\).
   - The distance from \((a, r)\) to the line \(mx - y = 0\) is:
     \[
     \frac{|ma - r|}{\sqrt{m^2 + 1}} = r
     \]
   - Solving for \(a\):
     \[
     |ma - r| = r\sqrt{m^2 + 1}
     \]
     \[
     ma - r = r\sqrt{m^2 + 1} \quad \text{or} \quad ma - r = -r\sqrt{m^2 + 1}
     \]
     \[
     ma = r(1 + \sqrt{m^2 + 1}) \quad \text{or} \quad ma = r(1 - \sqrt{m^2 + 1})
     \]

4. **Determine the relationship between the radii and the given conditions:**
   - Let the radii of the circles be \(r_1\) and \(r_2\).
   - Given that \(r_1 \cdot r_2 = 68\).

5. **Use homothety to relate the circles:**
   - Consider the homothety centered at the origin \(O\) that maps one circle to the other.
   - Let the homothety ratio be \(k\), such that \(k \cdot r_1 = r_2\).
   - Since the circles intersect at \((9,6)\), the homothety maps this point to itself, implying that the centers of the circles lie on the line through the origin and \((9,6)\).

6. **Calculate the slope \(m\):**
   - The line through the origin and \((9,6)\) has a slope \(m = \frac{6}{9} = \frac{2}{3}\).

7. **Express \(m\) in the required form:**
   - Given \(m = \frac{a\sqrt{b}}{c}\), we have \(a = 2\), \(b = 1\), and \(c = 3\).

8. **Sum the values of \(a\), \(b\), and \(c\):**
   \[
   a + b + c = 2 + 1 + 3 = 6
   \]

The final answer is \(\boxed{6}\)