H.C.F of $\frac{2}{3}$ , $\frac{4}{5}$, and $\frac{6}{7}$ is :

H.C.F of $\frac{2}{3}$ , $\frac{4}{5}$, and $\frac{6}{7}$ is :
1). $\frac{48}{105}$
2). $\frac{2}{105}$
3). $\frac{1}{105}$
4). $\frac{24}{105}$

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