If $\vec{i}+\vec{j}-\vec{k}, -\vec{i}+2\vec{j}+\vec{k}, \vec{j}+2\vec{k}, 2\vec{i}-\vec{j}+2\vec{k}$ are the position vectors of four points $A, B, C, D$ respectively,then the shortest distance between the lines $AB$ and $CD$ is

If $\bar{a}=(2 \hat{i}+2 \hat{j}+3 \hat{k})$,$\bar{b}=(-\hat{i}+2 \hat{j}+\hat{k})$ and $\bar{c}=(3 \hat{i}+\hat{j})$ such that $(\bar{a}+\lambda \bar{b})$ is perpendicular to $\bar{c}$,then the value of $\lambda$ is

Let $ABC$ be a triangle such that $\overrightarrow{BC} = \overrightarrow{a}$,$\overrightarrow{CA} = \overrightarrow{b}$,$\overrightarrow{AB} = \overrightarrow{c}$,$|\overrightarrow{a}| = 6\sqrt{2}$,$|\overrightarrow{b}| = 2\sqrt{3}$,and $\overrightarrow{b} \cdot \overrightarrow{c} = 12$. Consider the statements:
$(S1): |(\overrightarrow{a} \times \overrightarrow{b}) + (\overrightarrow{c} \times \overrightarrow{b})| - |\overrightarrow{c}| = 6(2\sqrt{2} - 1)$
$(S2): \angle ABC = \cos^{-1}\left(\sqrt{\frac{2}{3}}\right)$.
Which of the following is true?

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

Let $\vec{a} = 2\hat{i} + \hat{k}$, $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$. 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 $\vec{r} \cdot \vec{c} = $

If $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{r}$ is a non-zero vector such that $p\vec{r} + (\vec{r} \cdot \vec{b})\vec{a} = \vec{c}$,then $\vec{r} = $