If $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = r\hat{i} + \hat{j} + (2r - 1)\hat{k}$ are three vectors such that $\vec{c}$ is parallel to the plane of $\vec{a}$ and $\vec{b}$,then $r$ is equal to

If $\bar{p}, \bar{q}$ and $\bar{r}$ are non-zero,non-coplanar vectors,then $[\bar{p}+\bar{q}-\bar{r} \quad \bar{p}-\bar{q} \quad \bar{q}-\bar{r}] = \_\_\_\_$

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:

$(a+b) \cdot(b+c) \times(a+b+c)$ is equal to

Let $v = 2i + j - k$ and $w = i + 3k$. If $u$ is any unit vector, then the maximum value of the scalar triple product $[u v w]$ is

Let $a=\hat{i}+2 \hat{j}-\hat{k}$ and $b=\hat{i}+\hat{j}+\hat{k}$. If $p$ is a unit vector such that $[a b p]$ is maximum, then $p=$