If $\vec{p}$ and $\vec{q}$ are unequal unit vectors such that $(\vec{p} - \vec{q}) \cdot ((2\vec{q} + \vec{p}) \times (3\vec{p} - \vec{q})) = |\vec{p} + \vec{q}|$,then the angle between $\vec{p}$ and $\vec{q}$ is:

If $a, b$ and $c$ are three non-coplanar vectors and $p, q$ and $r$ are vectors defined by $p=\frac{b \times c}{[a b c]}, q=\frac{c \times a}{[a b c]}, r=\frac{a \times b}{[a b c]}$,then $(a+b) \cdot p+(b+c) \cdot q+(c+a) \cdot r$ is

If $\vec a = 2\sin \theta \hat i - \hat j + 2\hat k$,$\vec b = 2\hat i + 2\sin \theta \hat j - \hat k$ and $\vec c = 4\hat i + \hat j + 4\cos^2 \theta \hat k$ are coplanar,then $\theta$ can be equal to

The lines $\overline{r}=\overline{a}+\lambda(\overline{b} \times \overline{c})$ and $\overline{r}=\overline{c}+\mu(\overline{a} \times \overline{b})$ will intersect if

If $a=2u+3v+7w$,$b=u+v-2w$ and $c=-u-2v-3w$,then $\left|\frac{[u, v, w]}{[a, b, c]}\right|(a+b+c) = $

Let $\overrightarrow{a}=\hat{i}-2 \hat{j}+3 \hat{k}$,$\overrightarrow{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\overrightarrow{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1) \hat{k}$. If $\overrightarrow{c}$ is parallel to the plane containing $\overrightarrow{a}$ and $\overrightarrow{b}$,then $\lambda$ is equal to