If the points whose position vectors are $3i - 2j - k,$ $2i + 3j - 4k,$ $-i + j + 2k,$ and $4i + 5j + \lambda k$ lie on a plane,then $\lambda = $

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 $.....$

If the points $A(2,1,-1), B(0,-1,0), C(4,0,4)$ and $D(2,0,x)$ are coplanar,then $x=$

The value of $m$,if the vectors $\hat{\imath}-\hat{\jmath}-6 \hat{k}$,$\hat{\imath}-3 \hat{\jmath}+4 \hat{k}$,and $2 \hat{\imath}-5 \hat{\jmath}+m \hat{k}$ are coplanar,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:

Let $\vec{OD} = \hat{i} + 2\hat{j} + 6\hat{k}$ and $\vec{CB} = -3\hat{i} - 2\hat{k}$ be the diagonals of the parallelogram $OBDC$. If $\vec{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$, then the volume of the parallelepiped determined by vectors $\vec{OA}, \vec{OB}$, and $\vec{OC}$ (in cubic units) is: