If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}+3\hat{k}$,then $(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b}) = $ . . . . . . .

Let $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ be vectors such that $\bar{a} \times \bar{b} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\bar{c} \times \bar{d} = 3\hat{i} + 2\hat{j} + \lambda\hat{k}$. If $\begin{vmatrix} \bar{a} \cdot \bar{c} & \bar{b} \cdot \bar{c} \\ \bar{a} \cdot \bar{d} & \bar{b} \cdot \bar{d} \end{vmatrix} = 0$,then find the value of $\lambda$.

In a right trapezium $ABCD$,the diagonals are perpendicular,and the ratio of the length of bases $AD : BC = 2 : 3$. Then the ratio of the length of diagonals is

If $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ are orthogonal,then the value of $\lambda$ is:

Let $\lambda \in \mathbb{Z}$,$\vec{a} = \lambda \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$. Let $\vec{c}$ be a vector such that $(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = \vec{0}$,$\vec{a} \cdot \vec{c} = -17$ and $\vec{b} \cdot \vec{c} = -20$. Then $|\vec{c} \times (\lambda \hat{i} + \hat{j} + \hat{k})|^2$ is equal to

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+\hat{k}$,and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ are three vectors,then the vector $\vec{r}$ in the plane of $\vec{a}$ and $\vec{b}$,whose projection on $\vec{c}$ is $\frac{1}{\sqrt{3}}$,is given by: