The system of linear equations
$$x + y + z = 2$$
$$2x + 3y + 2z = 5$$
$$2x + 3y + (a^2 - 1)z = a + 1$$

We have the three simultaneous linear equations

$$x + y + z = 2$$

$$2x + 3y + 2z = 5$$

$$2x + 3y + (a^2 - 1)z = a + 1$$

To test consistency, we write the augmented matrix of the system:

$$\begin{bmatrix} 1 & 1 & 1 & \big| & 2\\ 2 & 3 & 2 & \big| & 5\\ 2 & 3 & a^2 - 1 & \big| & a + 1 \end{bmatrix}$$

Now we perform elementary row operations. First, eliminate the $$x$$-term from the second and third rows by replacing $$R_2$$ and $$R_3$$ with $$R_2 - 2R_1$$ and $$R_3 - 2R_1$$ respectively.

Doing the calculation for the second row:

$$\begin{aligned} R_2 - 2R_1 &:&\; (2-2\cdot1,\; 3-2\cdot1,\; 2-2\cdot1\;|\; 5-2\cdot2)\\ &= (0,\;1,\;0\;|\;1) \end{aligned}$$

Doing it for the third row:

$$\begin{aligned} R_3 - 2R_1 &:&\; (2-2\cdot1,\; 3-2\cdot1,\; (a^2-1)-2\cdot1\;|\; (a+1)-2\cdot2)\\ &= (0,\;1,\; a^2-3\;|\; a-3) \end{aligned}$$

The matrix is now

$$\begin{bmatrix} 1 & 1 & 1 & \big| & 2\\ 0 & 1 & 0 & \big| & 1\\ 0 & 1 & a^2 - 3 & \big| & a - 3 \end{bmatrix}$$

Next, eliminate the $$y$$-term from the third row by replacing $$R_3$$ with $$R_3 - R_2$$:

$$\begin{aligned} R_3 - R_2 &:&\; (0,\;1-1,\; (a^2-3)-0\;|\; (a-3)-1)\\ &= (0,\;0,\; a^2-3\;|\; a-4) \end{aligned}$$

So the reduced matrix is

$$\begin{bmatrix} 1 & 1 & 1 & \big| & 2\\ 0 & 1 & 0 & \big| & 1\\ 0 & 0 & a^2 - 3 & \big| & a - 4 \end{bmatrix}$$

At this stage we analyse the third row:

$$0x + 0y + (a^2 - 3)z = a - 4$$

If the coefficient $$(a^2-3)$$ is non-zero, we can solve for $$z$$ uniquely and the system is consistent. However, if $$(a^2-3)=0$$ while the right-hand side is non-zero, we get $$0 = \text{non-zero}$$, an impossibility, rendering the system inconsistent.

Set the coefficient equal to zero:

$$a^2 - 3 = 0 \;\;\Longrightarrow\;\; a = \pm\sqrt{3}$$

For $$a = \sqrt{3}$$ the right-hand side of the third row becomes

$$a - 4 = \sqrt{3} - 4 \neq 0$$

giving the contradictory statement $$0 = \sqrt{3}-4$$. Hence the system is inconsistent when $$a=\sqrt{3}$$.

For completeness, note that $$a=-\sqrt{3}$$ also makes the coefficient zero, and the right-hand side becomes $$-\sqrt{3}-4\neq0$$, giving another contradiction. But among the options given, only $$a=\sqrt{3}$$ is listed.

When $$a = 4$$, on the other hand, the coefficient is $$(4^2-3)=13\neq0$$, so we can find $$z = \dfrac{4-4}{13}=0$$, then $$y=1$$ from the second row, and finally $$x=1$$ from the first row; thus the system has a unique solution and is perfectly consistent.

Therefore the only statement that holds from the options is that the system is inconsistent when $$a = \sqrt{3}$$.

Hence, the correct answer is Option A.