The straight line is shown below with the intercepts.

$M\left(\frac{7}{2\cos \theta }, \frac{7}{2\sin \theta }\right)\equiv \left(h, k\right)$

$\therefore \cos \theta =\frac{7}{2h} \sin \theta =\frac{7}{2k}$

$\because \theta \in \left(0, \frac{\pi }{2}\right)$

$\Rightarrow \frac{49}{4h^2}+\frac{49}{4k^2}=1 \therefore x$ and $y$ are positive

$\therefore$ Locus $=\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{49}$

Now $\left(\alpha , \frac{7\sqrt{3}}{3}\right)$ lies on curve

$\Rightarrow \frac{1}{{\alpha }^2}+\frac{9}{147}=\frac{4}{49}$

$\frac{1}{{\alpha }^2}=\frac{1}{49}$

$\therefore \alpha =\pm 7$

$\therefore \alpha =7$

Hence this is the required option.