Consider the vectors x=i+2 j+3 k, y=2 i+3 j+k, and z=3 i+j+2 k For two distinct positive real numbers α and β , define X=α x+β y-z, Y=α y+β z-x, and Z=α z+β x-y If the vectors X, Y , and Z lie in a plane, the value of α+β-3 is ______ .

Correct answer: -2. ⇒ | α β -1 -1 α β β -1 α | underbrace | 1 2 3 2 3 1 3 1 2 |_-0=0 ⇒ (α^3+β^3-1 )-(-α β-α β-α β)=0 ⇒ α^3+β^3+3 α β=1 ⇒ α^3+β^3+(-1)^3=3(α)(β)(-1) ⇒ α+β-1=0 So, α+β-3=-2

Q28 / 48 · JEE Advanced 2025 (Paper 2) · Mathematics

Consider the vectors $\vec{x}=\hat{i}+2 \hat{j}+3 \hat{k}, \quad \vec{y}=2 \hat{i}+3 \hat{j}+\hat{k}, \quad \text { and } \quad \vec{z}=3 \hat{i}+\hat{j}+2 \hat{k}$ For two distinct positive real numbers $\alpha$ and $\beta$, define $\vec{X}=\alpha \vec{x}+\beta \vec{y}-\vec{z}, \quad \vec{Y}=\alpha \vec{y}+\beta \vec{z}-\vec{x}, \quad \text { and } \quad \vec{Z}=\alpha \vec{z}+\beta \vec{x}-\vec{y}$ If the vectors $\vec{X}, \vec{Y}$, and $\vec{Z}$ lie in a plane, the value of $\alpha+\beta-3$ is ______ .

Correct answer: -2

Solution

$\Rightarrow\left|\begin{array}{ccc}\alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha\end{array}\right| \underbrace{\left|\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2\end{array}\right|}_{-0}=0$ $\begin{aligned} & \Rightarrow\left(\alpha^3+\beta^3-1\right)-(-\alpha \beta-\alpha \beta-\alpha \beta)=0 \\ & \Rightarrow \alpha^3+\beta^3+3 \alpha \beta=1 \\ & \Rightarrow \alpha^3+\beta^3+(-1)^3=3(\alpha)(\beta)(-1) \\ & \Rightarrow \alpha+\beta-1=0\end{aligned}$ So, $\alpha+\beta-3=-2$

Hence, the correct answer is -2.

Solution

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