1. Find the intercepts of the line with the coordinate axes

   The family of lines is $x\cos\theta + y\sin\theta = 7, \qquad \theta\in\left(0,\frac{\pi}{2}\right).$

   • On the $x$-axis, $y=0$: $x\cos\theta=7 \implies x=\frac{7}{\cos\theta}=7\sec\theta.$ So the $x$-intercept is $\left(7\sec\theta,0\right)$.

   • On the $y$-axis, $x=0$: $y\sin\theta=7 \implies y=\frac{7}{\sin\theta}=7\csc\theta.$ So the $y$-intercept is $\left(0,7\csc\theta\right)$.

2. Find the midpoint of the intercept segment

   The midpoint of the segment joining $\left(7\sec\theta,0\right)$ and $\left(0,7\csc\theta\right)$ is $\left(\frac{7\sec\theta}{2},\frac{7\csc\theta}{2}\right).$

   Let this midpoint be $(x,y)$. Then $x=\frac{7}{2}\sec\theta, \qquad y=\frac{7}{2}\csc\theta.$

3. Use the given $y$-coordinate

   We are told that the point $\left(\alpha,\frac{7\sqrt3}{3}\right)$ lies on this locus. Hence, $\frac{7}{2}\csc\theta = \frac{7\sqrt3}{3}.$

   Cancelling $7$, $\frac{1}{2}\csc\theta = \frac{\sqrt3}{3}$ $\csc\theta = \frac{2\sqrt3}{3}.$

   Therefore, $\sin\theta = \frac{\sqrt3}{2}.$

   Since $\theta\in\left(0,\frac{\pi}{2}\right)$, $\theta=\frac{\pi}{3}.$

4. Now find $\alpha$

   $\alpha = x = \frac{7}{2}\sec\theta = \frac{7}{2}\sec\frac{\pi}{3} = \frac{7}{2}\cdot 2 = 7.$

5. Check options

   • A: $-7$ ✗
   • B: $7$ ✓
   • C: $-7\sqrt3$ ✗
   • D: $7\sqrt3$ ✗

Hence, the correct answer is B.