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Edward from Graveney School, Tooting, London sent this solution.
Let us make $a$ the radius of the largest circle centre $A$ etc. Then the lengths of the sides of the triangle are: $AB = a - b$, $AC = a - c$ and $BC = b + c$.
The perimeter of the triangle is: $$AB + BC + CA = (a - b) + (a - c) + (b + c)= 2a.$$ So the perimeter of the triangle is twice the radius of the large circle whatever the sizes of the small circles.