Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-coplanar unit vectors such that the angle between every pair of them is $\frac{\pi}{3}$. If $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} = p \vec{a} + q \vec{b} + r \vec{c}$,where $p, q$ and $r$ are scalars,then the value of $\frac{p^2 + 2q^2 + r^2}{q^2}$ is

If the volume of the parallelepiped formed by three non-coplanar vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is $4$ cubic units,then $[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}]$ is equal to

Let $\vec{a}=\hat{i}-\alpha \hat{j}+\beta \hat{k}$,$\vec{b}=3 \hat{i}+\beta \hat{j}-\alpha \hat{k}$ and $\vec{c}=-\alpha \hat{i}-2 \hat{j}+\hat{k}$,where $\alpha$ and $\beta$ are integers. If $\vec{a} \cdot \vec{b}=-1$ and $\vec{b} \cdot \vec{c}=10$,then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to $.....$

Let $\overrightarrow{A} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{B} = \hat{i}$,and $\overrightarrow{C} = C_1\hat{i} + C_2\hat{j} + C_3\hat{k}$. If $C_2 = -1$ and $C_3 = 1$,then to make the three vectors coplanar:

If the volume of a tetrahedron,whose vertices are with position vectors $\hat{i}-6 \hat{j}+10 \hat{k}$,$-\hat{i}-3 \hat{j}+7 \hat{k}$,$5 \hat{i}-\hat{j}+\lambda \hat{k}$ and $7 \hat{i}-4 \hat{j}+7 \hat{k}$ is $11$ cubic units,then the value of $\lambda$ is:

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector,assuming the question implies the vectors are linearly dependent or part of a coplanar set,find $\lambda$. Given the standard form of such problems,if we consider the vectors $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ to be coplanar with a reference vector,let us assume the third vector is $\hat{k}$. For these to be coplanar,the scalar triple product must be zero: $\left|\begin{array}{ccc} 1 & -3 & 1 \\ \lambda & 3 & 0 \\ 0 & 0 & 1 \end{array}\right| = 0$. Solving this,$\lambda$ is equal to: