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(D) Since $\vec{a}, \vec{b},$ and $\vec{c}$ are coplanar,their scalar triple product is zero:$[\vec{a} \vec{b} \vec{c}] = 0$$\begin{vmatrix} \alpha &

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$\alpha = 1, \beta = 1$

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(D) Since $\vec{a}, \vec{b},$ and $\vec{c}$ are coplanar,their scalar triple product is zero:$[\vec{a} \vec{b} \vec{c}] = 0$$\begin{vmatrix} \alpha & 2 & \beta \ 1 & 1 & 0 \ 0 & 1 & 1 \end{vmatrix} = 0$$\alpha(1-0) - 2(1-0) + \beta(1-0) = 0 \Rightarrow \alpha - 2 + \beta = 0 \Rightarrow \alpha + \beta = 2 \dots (i)$Since $\vec{a}$ bisects the angle between $\vec{b}$ and $\vec{c}$,$\vec{a}$ must be proportional to the sum of the unit vectors along $\vec{b}$ and $\vec{c}$:$\hat{b} = \frac{\vec{b}}{|\vec{b}|} = \frac{(1, 1, 0)}{\sqrt{2}}, \hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{(0, 1, 1)}{\sqrt{2}}$$\vec{a} = k(\hat{b} + \hat{c}) = \frac{k}{\sqrt{2}}(1, 2, 1)$Comparing this with $\vec{a} = (\alpha, 2, \beta)$,we get $\frac{k}{\sqrt{2}} = 1$,so $k = \sqrt{2}$.Thus,$\alpha = 1$ and $\beta = 1$.

The vector $\vec{a} = (\alpha, 2, \beta)$ lies in the plane of the vectors $\vec{b} = (1, 1, 0)$ and $\vec{c} = (0, 1, 1)$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which one of the following gives the possible values of $\alpha$ and $\beta$?

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