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

$ [\vec{a}+2 \vec{b}-\vec{c}, \vec{a}-\vec{b}, \vec{a}-\vec{b}-\vec{c}] $

Consider the vectors $\vec{a}=2 \hat{i}+3 \hat{j}-6 \hat{k}$, $\vec{b}=6 \hat{i}-2 \hat{j}+3 \hat{k}$ and $\vec{c}=3 \hat{i}-6 \hat{j}-2 \hat{k}$.
Assertion $(A):$ The three vectors do not form a triangle.
Reason $(R):$ The three vectors are non-coplanar.
The correct option among the following is:

If $\vec{u} = \hat{i} + 2\hat{j} - 2\hat{k}$, $\vec{v} = 2\hat{i} + \hat{k}$ and $\vec{w}$ is a unit vector, then the maximum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is

If $\overrightarrow{a} \cdot \overrightarrow{b} = 1, \overrightarrow{b} \cdot \overrightarrow{c} = 2$ and $\overrightarrow{c} \cdot \overrightarrow{a} = 3$,then the value of $[\vec{a} \times(\vec{b} \times \vec{c}), \vec{b} \times(\vec{c} \times \vec{a}), \vec{c} \times(\vec{b} \times \vec{a})]$ 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: