JEE Advance - Mathematics (2025 - Paper 2 Online - No. 12)
Consider the vectors
$$ \vec{x}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k}, \quad \vec{y}=2 \hat{\imath}+3 \hat{\jmath}+\hat{k}, \quad \text { and } \quad \vec{z}=3 \hat{\imath}+\hat{\jmath}+2 \hat{k} $$
For two distinct positive real numbers $\alpha$ and $\beta$, define
$$ \vec{X}=\alpha \vec{x}+\beta \vec{y}-\vec{z}, \quad \vec{Y}=\alpha \vec{y}+\beta \vec{z}-\vec{x}, \quad \text { and } \quad \vec{Z}=\alpha \vec{z}+\beta \vec{x}-\vec{y} . $$
If the vectors $\vec{X}, \vec{Y}$, and $\vec{Z}$ lie in a plane, then the value of $\alpha+\beta-3$ is ____________.
Explanation
$$ \begin{aligned} & {[\vec{x} \vec{y} \vec{z}]=0} \\ & \Rightarrow\left|\begin{array}{ccc} \alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha \end{array}\right| \underbrace{\left|\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{array}\right|}_{\neq 0}=0 \\ & \Rightarrow\left(\alpha^3+\beta^3-1\right)-(-\alpha \beta-\alpha \beta-\alpha \beta)=0 \\ & \Rightarrow \alpha^3+\beta^3+3 \alpha \beta=1 \\ & \Rightarrow \alpha^3+\beta^3+(-1)^3=3(\alpha)(\beta)(-1) \\ & \Rightarrow \alpha+\beta-1=0 \end{aligned} $$
So, $\alpha+\beta-3=-2$
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