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 $2 \hat{i}-\hat{j}+3 \hat{k}$, $-12 \hat{i}-\hat{j}-3 \hat{k}$, $-\hat{i}+2 \hat{j}-4 \hat{k}$ and $\lambda \hat{i}+2 \hat{j}-\hat{k}$ are the position vectors of four coplanar points, then $\lambda=$

If the points having the position vectors $-\hat{i}+4 \hat{j}-4 \hat{k}$,$3 \hat{i}+2 \hat{j}-5 \hat{k}$,$-3 \hat{i}+8 \hat{j}-5 \hat{k}$ and $-3 \hat{i}+2 \hat{j}+\lambda \hat{k}$ are coplanar,then $\lambda=$

If $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{b} = \hat{i} + 3\hat{k}$ and $\vec{c}$ is a unit vector, then the maximum value of the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is

Let $\vec{a} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k})$, $\vec{b} = (b_1\hat{i} + b_2\hat{j} + b_3\hat{k})$, and $\vec{c} = (c_1\hat{i} + c_2\hat{j} + c_3\hat{k})$ be three non-zero vectors such that $\vec{a}$ is a unit vector perpendicular to both $\vec{b}$ and $\vec{c}$. If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$, then find the value of $\left| \begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{matrix} \right|^2$.

Let the vectors $\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}$,$\vec{b}=(1-t) \hat{i}+(1+t) \hat{j}+2 \hat{k}$ and $\vec{c}=\hat{i}-t \hat{j}+\hat{k}$,$t \in R$ be such that for $\alpha, \beta, \gamma \in R$,$\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\vec{0} \Rightarrow \alpha=\beta=\gamma=0$. Then,the set of all values of $t$ is: