Scalar Triple Product

The value of $\lambda$ such that the four points whose position vectors are $3\vec i-2\vec j+\lambda\vec k,\ 6\vec i+3\vec j+\vec k,\ 5\vec i+7\vec j+3\vec k$ and $2\vec i+2\vec j+6\vec k$ are coplanar is

The value of $\lambda$ for which the volume of the parallelepiped formed by the vectors $\vec i+\lambda\vec j+\vec k,\ \vec j+\lambda\vec k,\ \lambda\vec i+\vec k$ is minimum is

$ \vec{v} = 2\hat{i} + \hat{j} - \hat{k},\quad \vec{w} = \hat{i} + 3\hat{k} $ If $ \vec{u} $ is a unit vector, maximum value of $ [\vec{u}\ \vec{v}\ \vec{w}] $ is:

If $ a,b,c $ are coplanar, evaluate $ [,2a - b, 2b - c,2c - a,] $

If there vectors $2\hat{i}-\hat{j}+\hat{k}$ , $\hat{i}+2\hat{j}-3\hat{k}$ and $3\hat{i}+\lambda \hat{j}+5\hat{k}$ are coplanar, then $\lambda$ is

If the vector $2\hat{i}-3\hat{j}$ , $\hat{i}+\hat{j}-\hat{k}$ and $3\hat{i}-\hat{k}$ form three conterminous edges of a parallelepiped, then thevolume of parallelepiped is

If $\vec{a}=\hat{i}-\hat{k},\, \vec{b}=x\hat{i}+\hat{j}+(1-x)\hat{k}$ and $\vec{c}=y\hat{i}+x\hat{j}+(1+x-y)\hat{k}$ , then $[\vec{a} , \vec{b}, \vec{c}]$ depends on

If $\vec{a}$ and $\vec{b}$ in space, given by $\vec{a}=\frac{\hat{i}-2\hat{j}}{\sqrt{5}}$ and $\vec{b}=\frac{2\hat{i}+\hat{j}+3\hat{k}}{\sqrt{14}}$ , then the value of $(2\vec{a}+\vec{b}).[(\vec{a} \times \vec{b}) \times (\vec{a}-2\vec{b})]$ is