Let $\vec{a}=3 \hat{i}+\hat{j}$ and $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$. Let $\vec{c}$ be a vector satisfying $\vec{a} \times(\vec{b} \times \vec{c})=\vec{b}+\lambda \vec{c}$. If $\vec{b}$ and $\vec{c}$ are non-parallel,then the value of $\lambda$ is.

If $\bar{a}=\frac{1}{\sqrt{10}}(3 \hat{\imath}+\hat{k})$ and $\bar{b}=\frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}-6 \hat{k})$,then the value of $(2 \bar{a}-\bar{b}) \cdot[(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b})]$ is

Let $\vec{a}$ be a non-zero vector. If $\vec{x}=\hat{i} \times(\vec{a} \times \hat{i})$, $\vec{y}=\hat{j} \times(\vec{a} \times \hat{j})-\vec{a}$ and $\vec{z}=\hat{k} \times(\vec{a} \times \hat{k})-\vec{a}$, then $\left[\begin{array}{lll}\vec{x} & \vec{y} & \vec{z}\end{array}\right]=$

If $\bar{x} \cdot \bar{y} = 0$,then $\bar{x} \times (\bar{x} \times \bar{y}) = \dots$ where $|\bar{x}| = 1$.

If $a = i + j - k$,$b = i - j + k$,and $c = i - j - k$,then $a \times (b \times c) = \dots$

If $\vec{a}$ and $\vec{b}$ are two non-collinear vectors, then $\frac{\vec{a} \times(\vec{b} \times \vec{a})}{|\vec{a}|^2}$ represents: