If the vectors $\overrightarrow{p}=(a+1) \hat{i}+a \hat{j}+a \hat{k}$,$\overrightarrow{q}=a \hat{i}+(a+1) \hat{j}+a \hat{k}$,and $\overrightarrow{r}=a \hat{i}+a \hat{j}+(a+1) \hat{k}$ $(a \in R)$ are coplanar and $3(\overrightarrow{p} \cdot \overrightarrow{q})^{2}-\lambda|\overrightarrow{r} \times \overrightarrow{q}|^{2}=0$,then the value of $\lambda$ is:

If $a = 3i - 2j + 2k$,$b = 6i + 4j - 2k$ and $c = 3i - 2j - 4k$,then $a \cdot (b \times c)$ 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 scalar $\overline{a} \cdot [(\overline{b} + \overline{c}) \times (\overline{a} + \overline{b} + \overline{c})]$ equals

Let $a = i - k$, $b = xi + j + (1 - x)k$, and $c = yi + xj + (1 + x - y)k$. Then $[a\,b\,c]$ depends on

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero non-coplanar vectors. Let the position vectors of four points $A, B, C$ and $D$ be $\vec{a}-\vec{b}+\vec{c}$,$\lambda \vec{a}-3 \vec{b}+4 \vec{c}$,$-\vec{a}+2 \vec{b}-3 \vec{c}$ and $2 \vec{a}-4 \vec{b}+6 \vec{c}$ respectively. If $\overrightarrow{AB}$,$\overrightarrow{AC}$ and $\overrightarrow{AD}$ are coplanar,then $\lambda$ is :