The vector $$\vec{a} = \alpha \hat{i} + 2\hat{j} + \beta \hat{k}$$ lies in the plane of the vectors $$\vec{b} = \hat{i} + \hat{j}$$ and $$\vec{c} = \hat{j} + \hat{k}$$ and bisects the angle between $$\vec{b}$$ and $$\vec{c}$$. Then which one of the following gives possible values of $$\alpha$$ and $$\beta$$?

The two given vectors are
$$\vec{b}= \hat{i}+ \hat{j}, \qquad \vec{c}= \hat{j}+ \hat{k}.$$

Their magnitudes are identical:
$$|\vec{b}|=\sqrt{1^{2}+1^{2}}=\sqrt{2}, \qquad |\vec{c}|=\sqrt{1^{2}+1^{2}}=\sqrt{2}.$$

For a vector to bisect the angle between $$\vec{b}$$ and $$\vec{c}$$, a well-known result from vector geometry states that the bisector must be proportional to the sum of the corresponding unit vectors. Hence
$$\vec{a}\ \parallel\ \frac{\vec{b}}{|\vec{b}|} + \frac{\vec{c}}{|\vec{c}|}.$$

Because $$|\vec{b}|=|\vec{c}|=\sqrt{2},$$ we obtain
$$\frac{\vec{b}}{|\vec{b}|} + \frac{\vec{c}}{|\vec{c}|} =\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})+\frac{1}{\sqrt{2}}(\hat{j}+\hat{k}) =\frac{1}{\sqrt{2}}\bigl(\hat{i}+2\hat{j}+\hat{k}\bigr).$$

Thus every angle-bisecting vector in the plane of $$\vec{b}$$ and $$\vec{c}$$ is a scalar multiple of
$$(\hat{i}+2\hat{j}+\hat{k}).$$

The problem specifies
$$\vec{a}=\alpha\hat{i}+2\hat{j}+\beta\hat{k},$$ whose $$\hat{j}$$-component is already fixed at $$2$$. Matching this with the candidate bisector $$k\,(\hat{i}+2\hat{j}+\hat{k})$$ forces the scale factor $$k$$ to be $$1$$, giving immediately

$$\alpha = 1,\qquad \beta = 1.$$

Therefore the only option that satisfies all conditions is:
Option D which is: $$\alpha = 1,\ \beta = 1.$$