1. Let the angles that the line makes with the positive coordinate axes be $\alpha,\beta,\gamma$ with the positive $x,y,z$-axes respectively.

   Given: $\beta=\frac{\alpha}{2},\qquad \gamma=\frac{\alpha}{2}$ Hence, $\alpha=2\beta,\qquad \gamma=\beta$

2. For direction angles of a line in 3D, the direction cosines satisfy $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1.$

   Substituting $\alpha=2\beta$ and $\gamma=\beta$: $\cos^2(2\beta)+\cos^2\beta+\cos^2\beta=1$ $\cos^2(2\beta)+2\cos^2\beta=1.$

3. Put $x=\cos^2\beta$. Then $\cos 2\beta=2\cos^2\beta-1=2x-1,$ so $\cos^2 2\beta=(2x-1)^2.$

   Therefore, $(2x-1)^2+2x=1$ $4x^2-4x+1+2x=1$ $4x^2-2x=0$ $2x(2x-1)=0.$

   So, $x=0 \quad \text{or} \quad x=\frac12.$

   That is, $\cos^2\beta=0 \quad \text{or} \quad \cos^2\beta=\frac12.$

4. Now find all possible values of $\beta$ in the range $[0,\pi]$ (direction angles lie in this interval):

   • If $\cos^2\beta=0$, then $\cos\beta=0 \implies \beta=\frac{\pi}{2}.$

   • If $\cos^2\beta=\frac12$, then $\cos\beta=\pm \frac{1}{\sqrt2} \implies \beta=\frac{\pi}{4},\frac{3\pi}{4}.$

   Hence all possible values are $\beta=\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4}.$

5. Their sum is

   =\frac{\pi+2\pi+3\pi}{4}=\frac{6\pi}{4}=\frac{3\pi}{2}. $$

6. Checking options:

   • A: $\frac{\pi}{2}$ — incorrect
   • B: $\pi$ — incorrect
   • C: $\frac{3\pi}{4}$ — incorrect
   • D: $\frac{3\pi}{2}$ — correct

Therefore, the correct answer is Option D.