If $a, b, c$ are non-coplanar vectors and $\lambda$ is a real number,then the vectors $a + 2b + 3c, \lambda b + 4c$ and $(2\lambda - 1)c$ are non-coplanar for

If the points $2a+3b-c, a-2b+3c, 3a+\lambda b-2c$ and $a-6b+6c$ are coplanar,then the direction cosines of the vector $\lambda \hat{i}-2\lambda \hat{j}+\hat{k}$ are

For any three non-zero vectors $\vec{r}_{1}, \vec{r}_{2}$ and $\vec{r}_{3}$,the determinant $\left| \begin{matrix} \vec{r}_{1} \cdot \vec{r}_{1} & \vec{r}_{1} \cdot \vec{r}_{2} & \vec{r}_{1} \cdot \vec{r}_{3} \\ \vec{r}_{2} \cdot \vec{r}_{1} & \vec{r}_{2} \cdot \vec{r}_{2} & \vec{r}_{2} \cdot \vec{r}_{3} \\ \vec{r}_{3} \cdot \vec{r}_{1} & \vec{r}_{3} \cdot \vec{r}_{2} & \vec{r}_{3} \cdot \vec{r}_{3} \end{matrix} \right| = 0$. Which of the following is false?

If $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$,$|\vec{a}|=2$,$|\bar{b}|=3$,$|\bar{c}|=4$ and the angle between $\bar{b}$ and $\bar{c}$ is $\frac{\pi}{3}$,then $\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]=$ (in $\sqrt{3}$)

$\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ 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:

The value of $\lambda$ for which points $A(2, 2, 1)$,$B(1, 1, 1)$,$C(-\lambda, 2, 1)$,and $D(3, 0, -1)$ are coplanar is $\lambda = $ ............