The maximum volume of a parallelepiped (in cubic units) with vectors $\vec{u} = 2a\hat{i} + \hat{k}$, $\vec{v} = a\hat{j} - a\hat{k}$, and $\vec{w} = 3\hat{i} + a\hat{j}$, where $a \in [0, 1]$, as its coterminous edges is:

If $\vec a = 2\sin \theta \hat i - \hat j + 2\hat k$,$\vec b = 2\hat i + 2\sin \theta \hat j - \hat k$ and $\vec c = 4\hat i + \hat j + 4\cos^2 \theta \hat k$ are coplanar,then $\theta$ can be equal to

If the four points $2\vec{a} + 3\vec{b} - \vec{c}$,$\vec{a} - 2\vec{b} + 3\vec{c}$,$3\vec{a} + 4\vec{b} - 2\vec{c}$,and $\vec{a} - \lambda\vec{b} - 6\vec{c}$ are coplanar,find the value of $\lambda$.

If $\vec{u} = \hat{j} + 4\hat{k}$,$\vec{v} = \hat{i} + 3\hat{k}$ and $\vec{w} = \cos \theta \hat{i} + \sin \theta \hat{j}$ are vectors in $3$-dimensional space,then the maximum possible value of $|(\vec{u} \times \vec{v}) \cdot \vec{w}|$ is

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

If the origin and the points $(1, 2, 3)$,$(2, 3, 4)$,and $(x, y, z)$ are coplanar,then