If a = i + j + k,b = 4i + 3j + 4k and c = i + | vedclass

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(D) Given that the vectors $a, b, c$ are linearly dependent,their scalar triple product must be zero: $[a, b, c] = 0$.This implies the determinant of

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$\alpha = \pm 1, \beta = 1$

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(D) Given that the vectors $a, b, c$ are linearly dependent,their scalar triple product must be zero: $[a, b, c] = 0$.This implies the determinant of the matrix formed by their components is zero:$\begin{vmatrix} 1 & 1 & 1 \ 4 & 3 & 4 \ 1 & \alpha & \beta \end{vmatrix} = 0$Expanding the determinant along the first row:$1(3\beta - 4\alpha) - 1(4\beta - 4) + 1(4\alpha - 3) = 0$$3\beta - 4\alpha - 4\beta + 4 + 4\alpha - 3 = 0$$-\beta + 1 = 0 \Rightarrow \beta = 1$.Given $|c| = \sqrt{3}$,we have $|c|^2 = 3$.$1^2 + \alpha^2 + \beta^2 = 3$$1 + \alpha^2 + 1^2 = 3$$\alpha^2 + 2 = 3 \Rightarrow \alpha^2 = 1 \Rightarrow \alpha = \pm 1$.Thus,$\alpha = \pm 1$ and $\beta = 1$.

If $a = i + j + k$,$b = 4i + 3j + 4k$ and $c = i + \alpha j + \beta k$ are linearly dependent vectors and $|c| = \sqrt{3}$,then

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