Consider the vectors

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

For two distinct positive real numbers $$\alpha$$ and $$\beta$$, define

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

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

The vectors lie in a plane if and only if their scalar triple product is zero, i.e. $$\vec{X}\cdot(\vec{Y}\times\vec{Z}) = 0$$. In component form this is equivalent to the determinant of the three column vectors being zero.

First write each required vector in $$\hat i,\hat j,\hat k$$ components.

$$\vec{x} = (1,\,2,\,3),\; \vec{y} = (2,\,3,\,1),\; \vec{z} = (3,\,1,\,2)$$

$$\vec{X}= \alpha\vec{x}+\beta\vec{y}-\vec{z} = (\alpha+2\beta-3,\; 2\alpha+3\beta-1,\; 3\alpha+\beta-2)$$

$$\vec{Y}= \alpha\vec{y}+\beta\vec{z}-\vec{x} = (2\alpha+3\beta-1,\; 3\alpha+\beta-2,\; \alpha+2\beta-3)$$

$$\vec{Z}= \alpha\vec{z}+\beta\vec{x}-\vec{y} = (3\alpha+\beta-2,\; \alpha+2\beta-3,\; 2\alpha+3\beta-1)$$

Introduce the shorthand

$$A = \alpha+2\beta-3,\qquad B = 2\alpha+3\beta-1,\qquad C = 3\alpha+\beta-2$$

Then$$ \vec{X} = (A,\;B,\;C),\; \vec{Y} = (B,\;C,\;A),\; \vec{Z} = (C,\;A,\;B). $$

Set up the determinant:

$$ \left| \begin{array}{ccc} A & B & C\\ B & C & A\\ C & A & B \end{array} \right| = 0. $$

Evaluating the determinant,

$$ \begin{aligned} D &= A(CB - A^{2}) - B(B^{2} - AC) + C(AB - C^{2}) \\ &= (ABC - A^{3}) + (ABC - B^{3}) + (ABC - C^{3}) \\ &= 3ABC - (A^{3}+B^{3}+C^{3}). \end{aligned} $$

Hence$$D = 0 \Longrightarrow A^{3}+B^{3}+C^{3}-3ABC = 0.$$ Using the identity$$ A^{3}+B^{3}+C^{3}-3ABC = (A+B+C)\,\bigl(A^{2}+B^{2}+C^{2}-AB-BC-CA\bigr), $$we get two possibilities:

$$A+B+C = 0.$$

$$A=B=C.$$

Because $$\alpha,\beta$$ are distinct positive numbers, Case 2 is impossible (solving $$A=B$$ forces $$\alpha+\beta=-2$$, which contradicts positivity). Therefore only Case 1 survives.

Compute $$A+B+C$$:

$$ \begin{aligned} A+B+C &= (\alpha+2\beta-3) + (2\alpha+3\beta-1) + (3\alpha+\beta-2)\\ &= 6\alpha + 6\beta - 6\\ &= 6(\alpha+\beta-1). \end{aligned} $$

Setting this to zero gives $$\alpha+\beta-1 = 0 \Longrightarrow \alpha+\beta = 1.$$ Both $$\alpha,\beta$$ can be positive and distinct while satisfying this relation, so the condition is admissible.

Finally, the required expression is

$$\alpha+\beta-3 = 1 - 3 = -2.$$

Hence the answer is -2.