If x is parallel to y and z where x = 2i + j | vedclass

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(C) Given that $x$ is parallel to both $y$ and $z$,it implies that $x$,$y$,and $z$ are coplanar. For three vectors to be coplanar,their scalar triple

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$ \pm \sqrt{7} $

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(C) Given that $x$ is parallel to both $y$ and $z$,it implies that $x$,$y$,and $z$ are coplanar. For three vectors to be coplanar,their scalar triple product must be zero,i.e.,$[x, y, z] = 0$. This is represented by the determinant: $\begin{vmatrix} 2 & 1 & \alpha \ \alpha & 0 & 1 \ 5 & -1 & 0 \end{vmatrix} = 0$ Expanding the determinant along the first row: $2(0(0) - 1(-1)) - 1(\alpha(0) - 1(5)) + \alpha(\alpha(-1) - 0(5)) = 0$ $2(1) - 1(-5) + \alpha(-\alpha) = 0$ $2 + 5 - \alpha^2 = 0$ $7 - \alpha^2 = 0$ $\alpha^2 = 7$ $\alpha = \pm \sqrt{7}$ Thus,the correct option is $C$.

If $x$ is parallel to $y$ and $z$ where $x = 2i + j + \alpha k$,$y = \alpha i + k$ and $z = 5i - j$,then $\alpha$ is equal to

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