1. Equation of the helicopter's path

The curve is $y - x^{3/2} = 7, \qquad x \ge 0$ so $y = x^{3/2} + 7.$

The soldier is at $\left(\frac12, 7\right).$

So a general point on the curve is $P(x, x^{3/2}+7), \quad x\ge 0.$

2. Distance from the soldier to a point on the curve

Distance squared is easier to minimize: $D^2 = \left(x-\frac12\right)^2 + \left((x^{3/2}+7)-7\right)^2.$

Thus, $D^2 = \left(x-\frac12\right)^2 + x^3.$

Let $f(x)=\left(x-\frac12\right)^2+x^3, \qquad x\ge 0.$ We minimize $f(x)$.

3. Differentiate and find critical points

$f(x)=x^2-x+\frac14+x^3$ so $f'(x)=2x-1+3x^2=3x^2+2x-1.$

Set $f'(x)=0$: $3x^2+2x-1=0.$

Solving, $x=\frac{-2\pm\sqrt{4+12}}{6}=\frac{-2\pm4}{6}.$

So, $x=\frac13 \quad \text{or} \quad x=-1.$

Since $x\ge 0$, only $x=\frac13$ is valid.

4. Check that it gives minimum

$f''(x)=6x+2.$ At $x=\frac13$, $f''\left(\frac13\right)=6\cdot\frac13+2=4>0.$ So this gives a minimum.

5. Compute the minimum distance

At $x=\frac13$, $D^2=\left(\frac13-\frac12\right)^2+\left(\frac13\right)^3.$

Now, $\frac13-\frac12=-\frac16 \implies \left(-\frac16\right)^2=\frac1{36},$ અને $\left(\frac13\right)^3=\frac1{27}.$

Hence, D^2=\frac1{36}+\frac1{27}=rac{3+4}{108}=rac7{108}.

Therefore, $D=\sqrt{\frac7{108}}=\frac{\sqrt7}{6\sqrt3}=\frac16\sqrt{\frac73}.$

6. Match with the options

$\boxed{D=\frac16\sqrt{\frac73}}$ which is Option A.

7. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.