Let $\vec{a} = -\hat{i} - \hat{k}$,$\vec{b} = -\hat{i} + \hat{j}$,and $\vec{c} = \hat{i} + 2\hat{j} + 3\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a} = 0$,then the value of $\vec{r} \cdot \vec{b}$ is

Let $\vec{AB} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{AD} = \hat{i} + 2 \hat{j} + \lambda \hat{k}$, $\lambda \in R$. Let the projection of the vector $\vec{v} = \hat{i} + \hat{j} + \hat{k}$ on the diagonal $\vec{AC}$ of the parallelogram $ABCD$ be of length $1$ unit. If $\alpha, \beta$, where $\alpha > \beta$, are the roots of the equation $\lambda^2 x^2 - 6 \lambda x + 5 = 0$, then $2 \alpha - \beta$ is equal to

Let three vectors $\overrightarrow{a}=\alpha \hat{i}+4 \hat{j}+2 \hat{k}$,$\vec{b}=5 \hat{i}+3 \hat{j}+4 \hat{k}$,and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ form a triangle such that $\overrightarrow{c}=\overrightarrow{a}-\overrightarrow{b}$ and the area of the triangle is $5 \sqrt{6}$. If $\alpha$ is a positive real number,then $|\overrightarrow{c}|^2$ is:

Let $a = i + 2j + k$, $b = i - j + k$, and $c = i + j - k$. $A$ vector in the plane of $a$ and $b$ has a projection of $\frac{1}{\sqrt{3}}$ on $c$. Then, one such vector is:

If $\bar{a}$ and $\bar{b}$ are not perpendicular to each other,$\bar{r} \times \bar{a} = \bar{b} \times \bar{a}$ and $\bar{r} \cdot \bar{c} = 0$,then $\bar{r} =$

In a triangle $ABC$, $D$ and $E$ divide the sides $BC$ and $CA$ in the ratio $2:1$ respectively. If $P$ is the point of intersection of $AD$ and $BE$, then the ratio in which $P$ divides $AD$ is