If $\vec{\alpha}=3 \vec{i}-\vec{k},|\vec{\beta}|=\sqrt{5}$ and $\vec{\alpha} \cdot \vec{\beta}=3$, then the area of the parallelogram for which $\vec{\alpha}$ and $\vec{\beta}$ are adjacent sides is

Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors. Suppose $\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}=0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{6}$. Then $\vec{a}$ is

A unit vector in XY-plane making an angle $$45^{\circ}$$ with $$\hat{i}+\hat{j}$$ and an angle $$60^{\circ}$$ with $$3 \hat{i}-4 \hat{j}$$ is

The value of 'a' for which the scalar triple product formed by the vectors $$\overrightarrow \alpha = \widehat i + a\widehat j + \widehat k,\overrightarrow \beta = \widehat j + a\widehat k$$ and $$\overrightarrow \gamma = a\widehat i + \widehat k$$ is maximum, is