$\vec{a}=\hat{i}+\hat{j}-2 \hat{k}$, $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}+\hat{j}-\hat{k}$ are three vectors. If $\vec{d}$ is a normal to the plane of $\vec{a}$ and $\vec{b}$ and $\vec{d} \cdot \vec{c}=2$, then $|\vec{d}|=$

For any three vectors $\vec{a}, \vec{b}, \vec{c}$,the value of $\vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times (\vec{c} + \vec{a}) + \vec{c} \times (\vec{a} + \vec{b})$ is equal to:

Let $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}$,$\vec{b}=3 \hat{i}+2 \hat{j}+5 \hat{k}$ and a vector $\vec{c}$ be such that $(\vec{a}-\vec{c}) \times \vec{b}=-18 \hat{i}-3 \hat{j}+12 \hat{k}$ and $\vec{a} \cdot \vec{c}=3$. If $\vec{b} \times \vec{c}=\vec{d}$,then $|\vec{a} \cdot \vec{d}|$ is equal to:

If the position vectors of the vertices of $\triangle ABC$ are $3\hat{i}+4\hat{j}-\hat{k}$, $\hat{i}+3\hat{j}+\hat{k}$, and $5(\hat{i}+\hat{j}+\hat{k})$ respectively, then the magnitude of the altitude from $A$ onto the side $BC$ is

Let $\vec{a} = \vec{j} - \vec{k}$ and $\vec{c} = \vec{i} - \vec{j} - \vec{k}$. Find the vector $\vec{b}$ satisfying $\vec{a} \times \vec{b} + \vec{c} = 0$ and $\vec{a} \cdot \vec{b} = 3$.

The vectors $a = xi + yj + zk$ and $b = j$ are such that $a, c, b$ form a right-handed system. Then $c$ is