If $3 \hat{i}-2 \hat{j}-\hat{k}$,$2 \hat{i}+3 \hat{j}-4 \hat{k}$,$-\hat{i}+\hat{j}+2 \hat{k}$ and $4 \hat{i}+5 \hat{j}+\lambda \hat{k}$ are respectively the position vectors of four coplanar points $P, Q, R$ and $S$,then $\lambda=$

Let $\overrightarrow{a}=\hat{i}+2\hat{j}+\hat{k}$ and $\overrightarrow{b}=3(\hat{i}-\hat{j}+\hat{k})$. Let $\overrightarrow{c}$ be a vector such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3$. Then $\overrightarrow{a} \cdot ((\overrightarrow{c} \times \overrightarrow{b})-\overrightarrow{b}-\overrightarrow{c})$ is equal to:

The volume of the tetrahedron whose coterminus edges are represented by $\bar{a}=-12 \hat{i}+p \hat{k}$,$\bar{b}=3 \hat{j}-\hat{k}$,and $\bar{c}=2 \hat{i}+\hat{j}-15 \hat{k}$ is $570$ cubic units. Then,$p=$

The number of integral values of $p$ for which the vectors $(p+1) \hat{i} - 3 \hat{j} + p \hat{k}$,$p \hat{i} + (p+1) \hat{j} - 3 \hat{k}$,and $-3 \hat{i} + p \hat{j} + (p+1) \hat{k}$ are linearly dependent is:

If the vectors $a \hat{i}+\hat{j}+\hat{k}$,$\hat{i}+b \hat{j}+\hat{k}$,and $\hat{i}+\hat{j}+c \hat{k}$ are coplanar,where $(a, b, c \neq 1)$,then the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$

$\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are non-coplanar vectors such that $\overrightarrow{P} = \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}$,$\overrightarrow{Q} = 4\overrightarrow{a} + 3\overrightarrow{b} + 4\overrightarrow{c}$,and $\overrightarrow{R} = \overrightarrow{a} + \alpha\overrightarrow{b} + \beta\overrightarrow{c}$ are linearly dependent vectors. Then,the number of possible values of $\alpha$ is: