Sum it

We know $(1+x)^{99}={99 \choose 0}+{99 \choose 1}x+{99 \choose 2}x^2+{99 \choose 3}x^3+.......+{99 \choose 99}x^{99}$

Put $x=1$ we get $\displaystyle \sum^{99}_{k=0}\dbinom{99}{k}=(1+1)^{99}=2^{99}={99 \choose 0}+{99 \choose 1}+{99 \choose 2}+{99 \choose 3}+.......+{99 \choose 99}$

Put $x=\omega$ then $(1+\omega)^{99}={99 \choose 0}+{99 \choose 1}\omega+{99 \choose 2}\omega^2+{99 \choose 3}.1+.......+{99 \choose 99}.1$

Now put $x=\omega^2$ then $(1+\omega^2)^{99}={99 \choose 0}+{99 \choose 1}\omega^2+{99 \choose 2}\omega+{99 \choose 3}.1+.......+{99 \choose 99}.1$

Adding these three equations we get:

$\displaystyle 3 \sum_{k=0}^{33} {99 \choose 3k}=2^{99} +(1+\omega)^{99} + (1+ \omega^2)^{99} \\ =2^{99} +(-\omega^2)^{99} +(-\omega)^{99} \\ =2^{99} -1-1=2^{99}-2 \\ \implies \displaystyle \sum_{k=0}^{33} {99 \choose 3k}=\dfrac{2^{99}-2}{3}$

After dividing $\dfrac{3.2^{99}}{2^{99}-2}=\dfrac{3}{2} \dfrac{2^{99}}{2^{98}-1}$

Note $\omega$ is the cube roots of unity and $1+\omega + \omega^2=0$

\\ \large\displaystyle \sum^{99}_{k=0}\dbinom{99}{k} = (1+1)^{99} = (2)^{99} \\ \large\displaystyle\sum^{33}_{k=0}\dbinom{99}{3k} = \dfrac{(1+1)^{99}+(1+\omega)^{99}+(1+\omega^2)^{99}}{3}, \text{ where } \omega \text{ is a third root of unity } \\ = \dfrac{2^{99}+(-\omega^2)^{99}+(-\omega)^{99}}{3} = \dfrac{(2^{99}-1-1)}{3} = \dfrac{(2^{99}-2)}{3} = \dfrac{2(2^{98}-1)}{3} \\ \therefore \large\displaystyle \sum^{99}_{k=0}\dbinom{99}{k} \text{divided by } \large\displaystyle\sum^{33}_{k=0}\dbinom{99}{3k} = \dfrac{3\times 2^{99}}{2\times (2^{98}-1)} = \large{\dfrac{3}{2}} \dot \large {\dfrac{2^{99}}{(2^{98}-1)}}. \\ \text{So } a = 3, b = 2, c = 2, d = 99, e = 2, f = 98 \implies 3+2+2+99+2+98-1 = \boxed{205}

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Note: We do not simplify the final term any further to preserve the form of the solution in the problem. $1$ is not a prime number.