Let $$A$$ be a $$3\times 3$$ real matrix such that

$$A\begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}1\\1\\0\end{pmatrix},\qquad A\begin{pmatrix}1\\0\\1\end{pmatrix}=\begin{pmatrix}-1\\0\\1\end{pmatrix},\qquad A\begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}1\\1\\2\end{pmatrix}.$$

If $$X=(x_1,x_2,x_3)^T$$ and $$I$$ is an identity matrix of order $$3$$, then the system

$$(A-2I)X=\begin{pmatrix}4\\1\\1\end{pmatrix}$$

has ____________.

Let

$$A=\begin{pmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{pmatrix}$$

Given,

$$A\begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}1\\1\\0\end{pmatrix}$$

Multiplying,

$$\begin{pmatrix}a_1+a_2\\b_1+b_2\\c_1+c_2\end{pmatrix}=\begin{pmatrix}1\\1\\0\end{pmatrix}$$

Hence,

$$a_1+a_2=1,\qquad b_1+b_2=1,\qquad c_1+c_2=0$$

Also,

$$A\begin{pmatrix}1\\0\\1\end{pmatrix}=\begin{pmatrix}-1\\0\\1\end{pmatrix}$$

Multiplying,

$$\begin{pmatrix}a_1+a_3\\b_1+b_3\\c_1+c_3\end{pmatrix}=\begin{pmatrix}-1\\0\\1\end{pmatrix}$$

Hence,

$$a_1+a_3=-1,\qquad b_1+b_3=0,\qquad c_1+c_3=1$$

Also,

$$A\begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}1\\1\\2\end{pmatrix}$$

Therefore,

$$a_3=1,\qquad b_3=1,\qquad c_3=2$$

Using these values,

$$a_1=-2,\qquad b_1=-1,\qquad c_1=-1$$

Now from

$$a_1+a_2=1,\qquad b_1+b_2=1,\qquad c_1+c_2=0$$

we get

$$a_2=3,\qquad b_2=2,\qquad c_2=1$$

Hence,

$$A=\begin{pmatrix}-2&3&1\\-1&2&1\\-1&1&2\end{pmatrix}$$

Now,

$$A-2I=\begin{pmatrix}-4&3&1\\-1&0&1\\-1&1&0\end{pmatrix}$$

The given system

$$(A-2I)\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}$$

gives

$$\begin{cases}-4x_1+3x_2+x_3=4\\-x_1+x_3=1\\-x_1+x_2=1\end{cases}$$

From the second equation,

$$x_3=x_1+1$$

From the third equation,

$$x_2=x_1+1$$

Substituting in the first equation,

$$-4x_1+3(x_1+1)+(x_1+1)=4$$

$$-4x_1+3x_1+3+x_1+1=4$$

$$4=4$$

Thus, the first equation is dependent on the other two equations.

Hence, one variable is free and the system has infinitely many solutions.

Therefore, the correct answer is $$\boxed{\text{infinitely many solutions}}$$.