If the vectors $\vec{a}=\lambda \hat{i}+\mu \hat{j}+4 \hat{k}$,$\vec{b}=2 \hat{i}+4 \hat{j}-2 \hat{k}$ and $\vec{c}=2 \hat{i}+3 \hat{j}+\hat{k}$ are coplanar and the projection of $\vec{a}$ on the vector $\vec{b}$ is $\sqrt{54}$ units,then the sum of all possible values of $\lambda+\mu$ is equal to:

Three vectors $\vec{a} = \hat{i} + \hat{j}$,$\vec{b} = \hat{j} + \hat{k}$,and $\vec{c} = \hat{k} + \hat{i}$ are given. If three unit vectors are drawn perpendicular to the three planes formed by these vectors,what is the volume of the parallelepiped formed by these unit vectors?

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors,then $\left| {\begin{array}{*{20}{c}} {\vec{a} \cdot \vec{a}} & {\vec{a} \cdot \vec{b}} & {\vec{a} \cdot \vec{c}} \\ {\vec{b} \cdot \vec{a}} & {\vec{b} \cdot \vec{b}} & {\vec{b} \cdot \vec{c}} \\ {\vec{c} \cdot \vec{a}} & {\vec{c} \cdot \vec{b}} & {\vec{c} \cdot \vec{c}} \end{array}} \right| = \dots$

If $a, b, c$ are any three non-coplanar vectors,then $[a + b, b + c, c + a] = $

Let $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{b}=\hat{j}-\hat{k}.$ If $\overrightarrow{c}$ is a vector such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3$,then $\vec{a} \cdot(\vec{b} \times \vec{c})$ is equal to :

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: