If the vectors $6i - 2j + 3k$,$2i + 3j - 6k$,and $3i + 6j - 2k$ are the position vectors of the vertices of a triangle,then the triangle is

If $a=\hat{i}+\hat{j}+\hat{k}$, $a \cdot b=1$ and $a \times b=\hat{j}-\hat{k}$, then $b=$

Let $\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}$,$\vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k}$,and $\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a}$ and $\vec{r} \cdot (\vec{b}-\vec{c})=0$,then $\frac{|593 \vec{r}+67 \vec{a}|^2}{(593)^2}$ is equal to:

The vector $2\hat{i} + a\hat{j} + \hat{k}$ is perpendicular to the vector $2\hat{i} - \hat{j} - \hat{k},$ if $a = $

Let $\vec{a}, \vec{b}, \vec{c}$ be non-coplanar vectors. If the three points with position vectors $\lambda \vec{a}-2 \vec{b}+\vec{c}$, $2 \vec{a}+\lambda \vec{b}-2 \vec{c}$, and $4 \vec{a}+7 \vec{b}-8 \vec{c}$ are collinear, then $\lambda=$

The angle $\theta$ between vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ is . . . . . . .