Consider the vectors vec{x}=hat{i}+2hat{j}+3h | vedclass

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(D) Since the vectors $\vec{X}, \vec{Y}, \vec{Z}$ lie in a plane,their scalar triple product must be zero,i.e.,$[\vec{X} \vec{Y} \vec{Z}] = 0$.This ca

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$-2$

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(D) Since the vectors $\vec{X}, \vec{Y}, \vec{Z}$ lie in a plane,their scalar triple product must be zero,i.e.,$[\vec{X} \vec{Y} \vec{Z}] = 0$.This can be written as the product of two determinants:$\begin{vmatrix} \alpha & \beta & -1 \ -1 & \alpha & \beta \ \beta & -1 & \alpha \end{vmatrix} \begin{vmatrix} 1 & 2 & 3 \ 2 & 3 & 1 \ 3 & 1 & 2 \end{vmatrix} = 0$.Calculating the second determinant: $1(6-1) - 2(4-3) + 3(2-9) = 5 - 2 - 21 = -18 
eq 0$.Thus,the first determinant must be zero: $\alpha^3 + \beta^3 + (-1)^3 - 3(\alpha)(\beta)(-1) = 0$.$\alpha^3 + \beta^3 - 1 + 3\alpha\beta = 0 \Rightarrow \alpha^3 + \beta^3 + (-1)^3 = 3\alpha\beta(-1)$.Using the identity $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$,we have $(\alpha+\beta-1)(\alpha^2+\beta^2+1-\alpha\beta+\alpha+\beta) = 0$.Since $\alpha, \beta > 0$,the second factor is always positive,so $\alpha+\beta-1 = 0$,which means $\alpha+\beta = 1$.Therefore,$\alpha+\beta-3 = 1-3 = -2$.

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