Consider the vectors \(\vec{x}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k}, \vec{y}=2 \hat{\imath}+3 \hat{\jmath}+\hat{k}\), a…

Since \(\vec{X}, \vec{Y}, \vec{Z}\) lie in a plane, their scalar triple product is \(0\). Hence the determinant formed by these three vectors must be \(0\).

Write \(\vec{X}, \vec{Y}, \vec{Z}\) in terms of \(\vec{x}, \vec{y}, \vec{z}\):

\(\vec{X}=\alpha\vec{x}+\beta\vec{y}-\vec{z},\quad \vec{Y}=-\vec{x}+\alpha\vec{y}+\beta\vec{z},\quad \vec{Z}=\beta\vec{x}-\vec{y}+\alpha\vec{z}\).

If \(B=[\vec{x}\ \vec{y}\ \vec{z}]\) and \(A=\begin{pmatrix}\alpha&-1&\beta\\ \beta&\alpha&-1\\ -1&\beta&\alpha\end{pmatrix}\), then \([\vec{X}\ \vec{Y}\ \vec{Z}]=BA\).

Therefore, \(\det[\vec{X}\ \vec{Y}\ \vec{Z}]=\det(B)\det(A)\).

Now,

\(\det(B)=\begin{vmatrix}1&2&3\\2&3&1\\3&1&2\end{vmatrix}=1(6-1)-2(4-3)+3(2-9)=-18\neq 0\).

So, for \(\vec{X}, \vec{Y}, \vec{Z}\) to be coplanar, we must have \(\det(A)=0\).

Now,

\(\det(A)=\begin{vmatrix}\alpha&-1&\beta\\ \beta&\alpha&-1\\ -1&\beta&\alpha\end{vmatrix}\)

\(=\alpha(\alpha^2+\beta)+(\alpha\beta-1)+\beta(\beta^2+\alpha)\)

\(=\alpha^3+\beta^3+3\alpha\beta-1\).

Hence,

\(\alpha^3+\beta^3+3\alpha\beta-1=0\).

Write it as

\(\alpha^3+\beta^3+(-1)^3-3\alpha\beta(-1)=0\).

Using \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\), with \(a=\alpha,\ b=\beta,\ c=-1\), we get

\((\alpha+\beta-1)(\alpha^2+\beta^2+1-\alpha\beta+\alpha+\beta)=0\).

Also,

\(\alpha^2+\beta^2+1-\alpha\beta+\alpha+\beta=\frac{1}{2}(\alpha-\beta)^2+\frac{1}{2}(\alpha+1)^2+\frac{1}{2}(\beta+1)^2>0\).

So the second factor cannot be \(0\). Therefore,

\(\alpha+\beta-1=0\Rightarrow \alpha+\beta=1\).

Hence,

\(\alpha+\beta-3=1-3=-2\).

Therefore, the correct answer is \(-2\).