The area of the quadrilateral $ABCD$ with vertices $A(0,4,1)$,$B(2,3,-1)$,$C(4,5,0)$,and $D(2,6,2)$ is equal to

If $a = (1, -1, 1)$ and $c = (-1, -1, 0),$ then the vector $b$ satisfying $a \times b = c$ and $a \cdot b = 1$ is

If non-zero vectors $a$ and $b$ are perpendicular to each other,then what is the solution for $r \times a = b$?

Let $L_1: \overrightarrow{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in R$,$L_2: \overrightarrow{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in R$,and $L_3: \overrightarrow{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k}), \delta \in R$ be three lines such that $L_1$ is perpendicular to $L_2$ and $L_3$ is perpendicular to both $L_1$ and $L_2$. Then the point which lies on $L_3$ is

Let $\vec{a}=\hat{i}-2\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ be two vectors. If $\vec{c}$ is a vector such that $\vec{b} \times \vec{c}=\vec{b} \times \vec{a}$ and $\vec{c} \cdot \vec{a}=0,$ then $\vec{c} \cdot \vec{b}$ is equal to

Let $\vec{u} = \hat{i} + \hat{j}$,$\vec{v} = \hat{i} - \hat{j}$,and $\vec{w} = \hat{i} + 2\hat{j} + 3\hat{k}$. If $\hat{n}$ is a unit vector such that $\vec{u} \cdot \hat{n} = 0$ and $\vec{v} \cdot \hat{n} = 0$,then $|\vec{w} \cdot \hat{n}| = ....$