If $\vec{x}$ is a unit vector such that $\vec{x} \times (\hat{i} - 2\hat{j} + \hat{k}) = -\hat{i} + \hat{k}$,then $\vec{x}$ is:

Let the vectors $\vec{a}, \vec{b}, \vec{c}$ be given as $\vec{a} = a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$,$\vec{b} = b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$,and $\vec{c} = c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}$. Then show that $\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$.

Let $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$. If $\vec c$ is a vector such that $\vec a \cdot \vec c + 2|\vec c| = 0$ and $|\vec c - \vec a| = \sqrt{14}$,and the angle between $\vec a \times \vec b$ and $\vec c$ is $30^o$,then $|(\vec a \times \vec b) \times \vec c|$ is:

If $a = i + 2j - 2k$,$b = 2i - j + k$ and $c = i + 3j - k$,then $a \times (b \times c)$ is equal to

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=\hat{i}-\hat{j}$ and if $6 \hat{i}+2 \hat{j}+3 \hat{k}=\lambda_1(\vec{a} \times \vec{b})+\lambda_2(\vec{b} \times \vec{c})+\lambda_3(\vec{c} \times \vec{a})$,then $(\lambda_1, \lambda_2, \lambda_3)=$

Let $\overrightarrow{a}=2 \hat{i}-\hat{j}+3 \hat{k}$,$\overrightarrow{b}=3 \hat{i}-5 \hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} \times \vec{c}=\vec{c} \times \vec{b}$ and $(\overrightarrow{a}+\overrightarrow{c}) \cdot(\overrightarrow{b}+\overrightarrow{c})=168$. Then the maximum value of $|\vec{c}|^2$ is :