Edward Wallace, age 18, 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: $\mathrm{AB}=a-b$, $\mathrm{AC}=a-c$ and $\mathrm{BC}=b+c$. The perimeter of the triangle is:

${\displaystyle \mathrm{AB}+\mathrm{BC}+\mathrm{CA}=(a-b)+(a-c)+(b+c)=2a.}$

So the perimeter of the triangle is twice the radius of the large circle whatever the sizes of the small circles.