The area of the triangle with vertices $(1,2,0)$,$(1,0,2)$ and $(0,3,1)$ is.

If $\theta$ is the angle between $\vec{f}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{g}=2 \hat{i}-3 \hat{j}+a \hat{k}$ and $\sin \theta=\sqrt{\frac{24}{28}}$,then $7 a^2+24 a=$

If the vertices of $\Delta ABC$ are $A=(2,3,5), B=(-1,3,2), C=(3,5,-2)$,then the area of the $\Delta ABC$ (in sq. units) is

$r \times a = b \times a;\,\,r \times b = a \times b;\,\,a \ne 0;\,\,b \ne 0;\,\,a \ne \lambda b;\,\,a$ is not perpendicular to $b,$ then $r = $

If $\overline{a}, \overline{b}, \overline{c}$ are three vectors such that $\overline{a} \neq \overline{0}$ and $\overline{a} \times \overline{b} = 2 \overline{a} \times \overline{c}$,$|\overline{a}| = |\overline{c}| = 1$,$|\overline{b}| = 4$ and $|\overline{b} \times \overline{c}| = \sqrt{15}$. If $\overline{b} - 2 \overline{c} = \lambda \overline{a}$,then $\lambda$ is

Let $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$, $\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}$ and a vector $\vec{c}$ be such that $2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0}$. If $\vec{a} \cdot \vec{c} = 15$, then $\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k})$ is equal to: