If one of the zeroes of the polynomial $3x^2-8x+\left(2k+1\right)$  is seven times the other, then $k=$

$\alpha ,7\alpha$

$\alpha +\beta =-\frac{b}{a}$

$\alpha +7\alpha =-\frac{\left(-8\right)}{3}$

$8\alpha =\frac{8}{3}$

$\alpha =\frac{1}{3}$

$\alpha \beta =\frac{c}{a}$

$7{\alpha }^2=\frac{2k+1}{3}$

$7{\left(\frac{1}{3}\right)}^2=\frac{2k+1}{3}$

$\frac{7}{9}=\frac{2k+1}{3}$

$7=3\left(2k+1\right)$

$7=6k+3$

$4=6k$

$\therefore k=\frac{2}{3}$