Each of the angles $$\beta$$ and $$\gamma$$ that a given line makes with the positive y- and z-axes, respectively, is half of the angle that this line makes with the positive x-axes. Then the sum of all possible values of the angle $$\beta$$ is

Let the direction angles of the line with the positive $$x$$-, $$y$$- and $$z$$-axes be $$\alpha , \beta , \gamma$$, respectively.
Their direction cosines are

$$l = \cos\alpha,\; m = \cos\beta,\; n = \cos\gamma$$

For every line in space, the direction cosines satisfy the identity

$$l^{2}+m^{2}+n^{2}=1 \;$$ $$-(1)$$

According to the question,

$$\beta = \dfrac{\alpha}{2},\quad \gamma = \dfrac{\alpha}{2}$$ $$-(2)$$

Substituting $$l,m,n$$ and relation $$(2)$$ into $$(1)$$ gives

$$\cos^{2}\alpha + \cos^{2}\!\left(\dfrac{\alpha}{2}\right) + \cos^{2}\!\left(\dfrac{\alpha}{2}\right) = 1$$

$$\Longrightarrow \cos^{2}\alpha + 2\cos^{2}\!\left(\dfrac{\alpha}{2}\right) = 1$$ $$-(3)$$

Use the half-angle identity $$\cos^{2}\!\left(\dfrac{\alpha}{2}\right)=\dfrac{1+\cos\alpha}{2}$$.
Substituting in $$(3)$$:

$$\cos^{2}\alpha + 2\left( \dfrac{1+\cos\alpha}{2} \right) = 1$$

$$\Longrightarrow \cos^{2}\alpha + 1 + \cos\alpha = 1$$

$$\Longrightarrow \cos^{2}\alpha + \cos\alpha = 0$$

Factorising:

$$\cos\alpha\,(\cos\alpha + 1) = 0$$

Therefore

$$\cos\alpha = 0 \quad \text{or} \quad \cos\alpha = -1$$

The direction angles lie in $$0 \le \alpha \le \pi$$.

If $$\cos\alpha = 0$$, then $$\alpha = \dfrac{\pi}{2}$$.
Using $$(2)$$, $$\beta = \dfrac{\alpha}{2} = \dfrac{\pi}{4}$$.

If $$\cos\alpha = -1$$, then $$\alpha = \pi$$.
Using $$(2)$$, $$\beta = \dfrac{\alpha}{2} = \dfrac{\pi}{2}$$.

Thus the possible values of $$\beta$$ are $$\dfrac{\pi}{4}$$ and $$\dfrac{\pi}{2}$$.

Required sum

$$\beta_{\text{sum}} = \dfrac{\pi}{4} + \dfrac{\pi}{2} = \dfrac{3\pi}{4}$$

Hence the sum of all possible values of $$\beta$$ is $$\dfrac{3\pi}{4}$$, corresponding to Option A.