Find a unit vector perpendicular to both vectors $3i + 2j - k$ and $12i + 5j - 5k$.

$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane defined by $\hat{i}$ and $\hat{i} + \hat{j}$,and the plane defined by $\hat{i} - \hat{j}$ and $\hat{i} + \hat{k}$. Find the angle between $\vec{a}$ and $\hat{i} - 2\hat{j} + 2\hat{k}$.

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 :

Find a unit vector perpendicular to each of the vectors $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b}),$ where $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}+3\hat{k}.$

Let $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}, \vec{c}=\hat{i}-\hat{j}$ and $\vec{d}=\hat{i}+\hat{j}+x \hat{k}$. If $(\vec{a} \times \vec{b}) \times \vec{c}$ is perpendicular to $\vec{d}$, then $x=$

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{b}|=1$ and $|\vec{b} \times \vec{a}|=2$. Then $|(\vec{b} \times \vec{a})-\vec{b}|^2$ is equal to