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Q.
If $a =\displaystyle\sum_{ r =0}^{33}{ }^{33} C _{ r }, b =\displaystyle\sum_{ r =0}^{33}{ }^{66} C _{2 r }, c =\displaystyle\sum_{ r =0}^{33}{ }^{99} C _{3 r }$, then
Binomial Theorem
Solution:
${ }^{ n } C _0+{ }^{ n } C _1+{ }^{ n } C _2 \ldots .{ }^{ n } C _{ n }=2^{ n } \Rightarrow a =2^{33}$
${ }^n C_0+{ }^n C_2+{ }^n C_4+\ldots=2^{n-1} \Rightarrow b=2^{65}$
$(1+x)^n={ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+{ }^n C_3 x^3+\ldots . .+{ }^n C_n x^n$
Put $x=1, \omega$ and $\omega^2$ and add them, we get
$C =\frac{2^{99}-2}{3}$
Now check the answer