If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j}$,$\vec{c} = \hat{i}$ and $(\vec{a} \times \vec{b}) \times \vec{c} = \lambda \vec{a} + \mu \vec{b}$,then $\lambda + \mu$ is equal to :-

If $\vec{a} = \hat{i} + 2\hat{j} - 2\hat{k}$,$\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$,and $\vec{c} = \hat{i} + 3\hat{j} - \hat{k}$,then find $\vec{a} \times (\vec{b} \times \vec{c})$.

If $(\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c})$ where $\vec{a}, \vec{b},$ and $\vec{c}$ are any three vectors such that $\vec{a} \cdot \vec{b} \neq 0$ and $\vec{b} \cdot \vec{c} \neq 0$,then $\vec{a}$ and $\vec{c}$ are:

If $a=(1,2,3), b=(2,-1,1), c=(3,2,1)$ and $a \times(b \times c)=\alpha a+\beta b+\gamma c$,then

Let $\vec{a}=-\hat{i}-\hat{j}+\hat{k}$,$\vec{a} \cdot \vec{b}=1$ and $\vec{a} \times \vec{b}=\hat{i}-\hat{j}$. Then $\vec{a}-6 \vec{b}$ is equal to

If $a = i + j - 2k$,then $\sum \{(a \times i) \times j\}^2$ is equal to