If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and $(\vec{a} - \lambda \vec{b}) \cdot (\vec{b} - 2\vec{c}) \times (\vec{c} + 2\vec{a}) = 0$,then $\lambda$ is equal to

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

$(\bar{a}+2 \bar{b}-\bar{c}) \cdot \{(\bar{a}-\bar{b}) \times (\bar{a}-\bar{b}-\bar{c})\} = $

If $|\vec{a}| = 4, |\vec{b}| = 3$ and $\vec{a} \cdot \vec{b} = 8$, then the scalar triple product $[\vec{a} \quad \vec{a} + \vec{b} \quad \vec{a} \times \vec{b}]$ is equal to:

If the points $2a+3b-c, a-2b+3c, 3a+\lambda b-2c$ and $a-6b+6c$ are coplanar, then the direction cosines of the vector $\lambda \hat{i}-2\lambda \hat{j}+\hat{k}$ are

The sum of the distinct real values of $\mu$,for which the vectors $\mu \hat{i} + \hat{j} + \hat{k}$,$\hat{i} + \mu \hat{j} + \hat{k}$,and $\hat{i} + \hat{j} + \mu \hat{k}$ are coplanar,is