$\text{Consider the vectors}$

$\overrightarrow{x}=\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{y}=2\hat{i}+3\hat{j}+\hat{k},\overrightarrow{z}=3\hat{i}+\hat{j}+2\hat{k}.$

$\text{For two distinct positive real numbers}\mathrm{ \alpha }\text{ and}\mathrm{ \beta },\text{define}$

$\overrightarrow{X}=\alpha \overrightarrow{x}+\beta \overrightarrow{y}-\overrightarrow{z},$

$\overrightarrow{Y}=\alpha \overrightarrow{y}+\beta \overrightarrow{z}-\overrightarrow{x},\text{and}$

$\overrightarrow{Z}=\alpha \overrightarrow{z}+\beta \overrightarrow{x}-\overrightarrow{y}.$

$\text{If the vectors}\overrightarrow{X},\overrightarrow{Y},\text{and}\overrightarrow{Z}\text{lie in a plane, then the value of}\alpha +\beta -3\text{is}\text{\_\_\_\_\_\_}$

The correct answer is -2.

Step-by-step Explanation:

We are given three vectors in three-dimensional space:
$\overrightarrow{x}=\hat{i}+2\hat{j}+3\hat{k}$
$\overrightarrow{y}=2\hat{i}+3\hat{j}+\hat{k}$
$\overrightarrow{z}=3\hat{i}+\hat{j}+2\hat{k}$

We are also given three other vectors defined as:
$\overrightarrow{X}=\alpha \overrightarrow{x}+\beta \overrightarrow{y}-\overrightarrow{z}$
$\overrightarrow{Y}=-\overrightarrow{x}+\alpha \overrightarrow{y}+\beta \overrightarrow{z}$
$\overrightarrow{Z}=\beta \overrightarrow{x}-\overrightarrow{y}+\alpha \overrightarrow{z}$

Since the vectors $\overrightarrow{X},\overrightarrow{Y},\text{and}\overrightarrow{Z}$ lie in a plane (i.e., they are coplanar), their scalar triple product must be equal to zero:
$\left[\overrightarrow{X}\overrightarrow{Y}\overrightarrow{Z}\right]=0$

Expressing the scalar triple product of $\overrightarrow{X},\overrightarrow{Y},\overrightarrow{Z}$ in terms of $\overrightarrow{x},\overrightarrow{y},\overrightarrow{z}$ , we have:
$\left[\overrightarrow{X}\overrightarrow{Y}\overrightarrow{Z}\right]=\left|\begin{array}{ccc}\alpha & \beta & -1\\ -1& \alpha & \beta \\ \beta & -1& \alpha \end{array}\right|\left[\overrightarrow{x}\overrightarrow{y}\overrightarrow{z}\right]$

First, let us check the scalar triple product $\left[\overrightarrow{x}\overrightarrow{y}\overrightarrow{z}\right]$ :
$\left[\overrightarrow{x}\overrightarrow{y}\overrightarrow{z}\right]=\left|\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\\ 3& 1& 2\end{array}\right|$
Evaluating this determinant:
$\left[\overrightarrow{x}\overrightarrow{y}\overrightarrow{z}\right]=1(6-1)-2(4-3)+3(2-9)=5-2-21=-18$
Since $\left[\overrightarrow{x}\overrightarrow{y}\overrightarrow{z}\right]\ne 0$ , the coefficient determinant must be equal to zero:
$\left|\begin{array}{ccc}\alpha & \beta & -1\\ -1& \alpha & \beta \\ \beta & -1& \alpha \end{array}\right|=0$

Expanding the determinant:
$\alpha ({\alpha }^2+\beta )-\beta (-\alpha -{\beta }^2)-1(1-\alpha \beta )=0$
Simplifying the terms:
${\alpha }^3+\alpha \beta +\alpha \beta +{\beta }^3-1+\alpha \beta =0$
${\alpha }^3+{\beta }^3-1+3\alpha \beta =0$

Using the algebraic identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ with $a=\alpha$ , $b=\beta$ , and $c=-1$ , we get:
$(\alpha +\beta -1)({\alpha }^2+{\beta }^2+1-\alpha \beta +\alpha +\beta )=0$

Since $\alpha$ and $\beta$ are positive real numbers, the expression ${\alpha }^2-\alpha \beta +{\beta }^2+\alpha +\beta +1$ is strictly positive and can never be zero. Therefore, we must have:
$\alpha +\beta -1=0\Rightarrow \alpha +\beta =1$

We need to find the value of $\alpha +\beta -3$ :
$\alpha +\beta -3=1-3=-2$