$[(a \times b) \times (b \times c), (b \times c) \times (c \times a), (c \times a) \times (a \times b)] = \,$

The vectors $i + 2j + 3k$,$\lambda i + 4j + 7k$,and $-3i - 2j - 5k$ are collinear if $\lambda$ equals:

Let $\vec{a} = \hat{i} - \hat{k}$,$\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}$,and $\vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$. Then the scalar triple product $[\vec{a} \, \vec{b} \, \vec{c}]$ depends on:

Three vectors $\hat{i}-\hat{k}$,$\lambda \hat{i}+\hat{j}+(1-\lambda) \hat{k}$,and $\mu \hat{i}+\lambda \hat{j}+(1+\lambda-\mu) \hat{k}$ represent the coterminous edges of a parallelepiped. The volume of the parallelepiped depends on:

The sum of all values of $\alpha$,for which the points whose position vectors $\hat{i}-2 \hat{j}+3 \hat{k}$,$2 \hat{i}-3 \hat{j}+4 \hat{k}$,$(\alpha+1) \hat{i}+2 \hat{k}$ and $9 \hat{i}+(\alpha-8) \hat{j}+6 \hat{k}$ are coplanar,is equal to

Let $p, q, r$ be three non-coplanar vectors and $b = p \times q$. If $a, b, c$ denote the coterminous edges of a parallelepiped,then its height with the base having $a$ and $c$ is