Let $\overrightarrow{a}=\hat{i}-2 \hat{j}+3 \hat{k}$, $\overrightarrow{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\overrightarrow{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1) \hat{k}$. If $\overrightarrow{c}$ is parallel to the plane containing $\overrightarrow{a}$ and $\overrightarrow{b}$, then $\lambda$ is equal to

If the vectors $\vec{a}=\lambda \hat{i}+\mu \hat{j}+4 \hat{k}$,$\vec{b}=2 \hat{i}+4 \hat{j}-2 \hat{k}$ and $\vec{c}=2 \hat{i}+3 \hat{j}+\hat{k}$ are coplanar and the projection of $\vec{a}$ on the vector $\vec{b}$ is $\sqrt{54}$ units,then the sum of all possible values of $\lambda+\mu$ 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

If $a, b, c$ are three non-coplanar vectors,then $\frac{a \cdot (b \times c)}{c \times a \cdot b} + \frac{b \cdot (a \times c)}{c \cdot (a \times b)} = $

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$,$\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$,and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$ be three non-zero vectors such that $\vec{c}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$. If the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,then $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|^2 = \dots$

If $a$ and $b$ are unit vectors perpendicular to each other, then $[a + (a \times b) \quad b + (a \times b) \quad (a \times b)]$ =