Refer to the attached image. `O_1 ` denotes the center of `Z , ` `O_2 ` denotes the center of `W . ` Let `R = 7 ` be the radius of `Z .`

First, because the angle QPR is equal to 60 degrees, its central angle `QO_2R ` is equal to 120 degrees. Second, draw the line `O_1O_2 ` until it intersects `QR ` at the point `S . ` From the symmetry, this line is a median of the triangle `QO_1R , ` so it is its height, too. And for the isosceles triangle `QO_2R ` it is also a bisector, so the angles `QO_2S ` and `RO_2S ` are equal to 60 degrees, while the angles `O_2QS ` and `O_2RS ` are equal to 30 degrees.

Because of this, the radius of W is `r = O_2Q = OS / cos ( 30^@ ) = 3 ` and `O_2S = r / 2 = 3 / 2 .`

From the right triangle `QSO_1 ` we know `SO_1 = sqrt(R^2-(3sqrt(3)/2)^2) = sqrt(49-27/4)=13/2.`

Now draw a line perpendicular to `O_1O_2 ` that goes through the point P; let it intersect `O_1O_2 ` at H. Then HS has the same length as the height of QPR from the point P, so the area in question is `1 / 2 QR * SH .`

To find SH, consider the right triangle `O_1PO_2 . ` Its hypotenuse is `O_1O_2 = 13/2 - 3/2 = 5, ` so `PO_1 = 4. ` Its height PH has the length `h = ( 3*4 ) / 5 ` from the area considerations. Then `O_1H ` is equal to `4 / 3 h = 16/5 ` from the similar triangles `O_1PO_2 ` and `O_1HP .`

Finally, `HS = 13/2 - 16/5 = 33/10 ` and we see that the area of QPR is

`1 / 2 QR * SH = (3 sqrt(3)) / 2 * 33/10 = (99sqrt(3)) /20.`

This way, `a=99, b=3, c=20 ` and `a+b+c=122.`

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