If $$\alpha + \beta + \gamma = 2\pi$$, then the system of equations
$$x + \cos\gamma y + \cos\beta z = 0$$
$$\cos\gamma x + y + \cos\alpha z = 0$$
$$\cos\beta x + \cos\alpha y + z = 0$$
has:

We are given the homogeneous linear system

$$$\begin{aligned} x+\cos\gamma\,y+\cos\beta\,z &= 0,\\ \cos\gamma\,x+y+\cos\alpha\,z &= 0,\\ \cos\beta\,x+\cos\alpha\,y+z &= 0, \end{aligned}$$$

together with the relation $$\alpha+\beta+\gamma = 2\pi.$$ For a homogeneous system, the number of non-trivial solutions is decided completely by the determinant of its coefficient matrix. If the determinant is non-zero, the only solution is the trivial one $$(x,y,z)=(0,0,0).$$ If the determinant is zero, the matrix is singular and we obtain infinitely many non-trivial solutions in addition to the trivial one.

We therefore form the coefficient matrix

$$$A=\begin{bmatrix} 1 & \cos\gamma & \cos\beta\\ \cos\gamma & 1 & \cos\alpha\\ \cos\beta & \cos\alpha & 1 \end{bmatrix}$$$

and compute its determinant. Using the 3 × 3 determinant expansion formula

$$$\det A = 1\bigl(1\cdot1-\cos^2\alpha\bigr) -\cos\gamma\bigl(\cos\gamma\cdot1-\cos\alpha\cos\beta\bigr) +\cos\beta\bigl(\cos\gamma\cos\alpha-1\cdot\cos\beta\bigr), $$$

we proceed step by step.

First term:

$$1\bigl(1\cdot1-\cos^2\alpha\bigr)=1-\cos^2\alpha.$$

Second term:

$$-\cos\gamma\bigl(\cos\gamma-\cos\alpha\cos\beta\bigr) =-\cos^2\gamma+\cos\alpha\cos\beta\cos\gamma.$$

Third term:

$$\cos\beta\bigl(\cos\gamma\cos\alpha-\cos\beta\bigr) =\cos\alpha\cos\beta\cos\gamma-\cos^2\beta.$$

Adding the three results we get

$$$\det A = 1-\cos^2\alpha-\cos^2\beta-\cos^2\gamma +2\cos\alpha\cos\beta\cos\gamma.$$$

Thus

$$$\boxed{\;\det A =1+2\cos\alpha\cos\beta\cos\gamma -\bigl(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\bigr)\;}.$$$

To decide the sign of this expression we now use the given condition $$\alpha+\beta+\gamma=2\pi.$$ Because cosine has period $$2\pi$$, we can write

$$\cos\alpha=\cos(2\pi-\beta-\gamma)=\cos(\beta+\gamma).$$

Applying the compound-angle formula $$\cos(P+Q)=\cos P\,\cos Q-\sin P\,\sin Q,$$ we find

$$\cos\alpha=\cos\beta\cos\gamma-\sin\beta\sin\gamma.$$ Denote $$$\cos\beta=B,\qquad\cos\gamma=C,\qquad \sin\beta=S,\qquad\sin\gamma=T.$$$ Then $$\cos\alpha=BC-ST.$$

We substitute these symbols into the determinant expression. The terms become

$$$\cos\alpha=BC-ST,\qquad \cos^2\alpha=(BC-ST)^2,\qquad \cos\beta=B,\qquad \cos\gamma=C.$$$

Hence

$$$\det A =1+2(BC-ST)BC-\bigl[(BC-ST)^2+B^2+C^2\bigr].$$$

Expanding step by step:

1. The mixed product term:

$$2(BC-ST)BC =2B^2C^2-2STBC.$$

2. The squared term:

$$(BC-ST)^2 =B^2C^2-2STBC+S^2T^2.$$

Putting everything together,

$$$\det A =1+2B^2C^2-2STBC -\bigl[B^2C^2-2STBC+S^2T^2+B^2+C^2\bigr].$$$

Now distribute the minus sign carefully:

$$$\det A =1+2B^2C^2-2STBC -B^2C^2+2STBC-S^2T^2-B^2-C^2.$$$

The terms $$-2STBC$$ and $$+2STBC$$ cancel out. Combining like terms we obtain

$$\det A =1+B^2C^2-S^2T^2-B^2-C^2.$$

Next we bring in the Pythagorean identity $$\sin^2\theta=1-\cos^2\theta.$$ Hence $$S^2=1-B^2,\qquad T^2=1-C^2,$$ so that

$$$S^2T^2=(1-B^2)(1-C^2)=1-B^2-C^2+B^2C^2.$$$

Replace $$S^2T^2$$ in the determinant:

$$$\det A =1+B^2C^2-\bigl[1-B^2-C^2+B^2C^2\bigr]-B^2-C^2.$$$ Distribute the minus sign once more:

$$$\det A =1+B^2C^2-1+B^2+C^2-B^2C^2-B^2-C^2.$$$

Immediately every term cancels:

$$\det A=0.$$

Because $$\det A=0,$$ the coefficient matrix is singular. A homogeneous linear system with a singular coefficient matrix always possesses infinitely many solutions (every scalar multiple of any non-trivial solution is also a solution, in addition to the trivial solution).

Hence, the correct answer is Option A.