If the points with position vectors $\hat{i}-\hat{j}+\hat{k}$, $2 \hat{i}-\hat{k}$, $\hat{j}+2 \hat{k}$ and $\hat{i}+\hat{j}+\lambda \hat{k}$ are coplanar, then the magnitude of the vector $6 \lambda \hat{i}-3 \hat{j}+6 \hat{k}$ is

If $a = 2i + j - k$,$b = i + 2j + k$,and $c = i - j + 2k$,then $a \cdot (b \times c) = \dots$

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

Let $a, b, c$ be three distinct real numbers,none equal to $1$. 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,then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is equal to

Let $\overrightarrow{a}$ be a unit vector, $\overrightarrow{b} = 2\hat{i} + \hat{j} - \hat{k}$ and $\overrightarrow{c} = \hat{i} + 3\hat{k}$. Then, the maximum value of $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$ is

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector, let us assume the vectors are $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$, $\vec{b} = \lambda \hat{i}+3 \hat{j}$, and we consider the standard basis vectors or a third vector to define coplanarity. However, if the question implies these two vectors are coplanar with the origin or a specific plane, we evaluate the scalar triple product. Given the standard interpretation of such problems, if $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ are coplanar with $\vec{c} = \hat{j}$, then the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 0$. Solving for $\lambda$ where $\vec{a} = (1, -3, 1)$, $\vec{b} = (\lambda, 3, 0)$, and $\vec{c} = (0, 1, 0)$: