If $\bar{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \bar{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$,and $\bar{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$,and $[3 \bar{a}+\bar{b} \quad 3 \bar{b}+\bar{c} \quad 3 \bar{c}+\bar{a}] = \lambda \begin{vmatrix} \bar{a} \cdot \hat{i} & \bar{a} \cdot \hat{j} & \bar{a} \cdot \hat{k} \\ \bar{b} \cdot \hat{i} & \bar{b} \cdot \hat{j} & \bar{b} \cdot \hat{k} \\ \bar{c} \cdot \hat{i} & \bar{c} \cdot \hat{j} & \bar{c} \cdot \hat{k} \end{vmatrix}$,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:

The volume of the tetrahedron whose co-terminus edges are $\bar{a}, \bar{b}, \bar{c}$ is $12$ cubic units. If the scalar projection of $\bar{a}$ on $\bar{b} \times \bar{c}$ is $4$,then $|\bar{b} \times \bar{c}|=$

If $a, b, c$ are non-coplanar vectors and $d = \lambda a + \mu b + \nu c$,then $\lambda = \dots$

If $\bar{x}=\frac{\bar{b} \times \bar{c}}{[\bar{a} \bar{b} \bar{c}]}, \bar{y}=\frac{\bar{c} \times \bar{a}}{[\bar{a} \bar{b} \bar{c}]}$ and $\bar{z}=\frac{\bar{a} \times \bar{b}}{[\bar{a} \bar{b} \bar{c}]}$ where $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors,then the value of $\bar{x} \cdot(\bar{a}+\bar{b})+\bar{y} \cdot(\bar{b}+\bar{c})+\bar{z} \cdot(\bar{c}+\bar{a})$ is

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