The value of $(\vec{a} + 2\vec{b} - \vec{c}) \cdot \{(\vec{a} - \vec{b}) \times (\vec{a} - \vec{b} - \vec{c})\}$ is equal to

If three vectors $\vec{a} = 12\hat{i} + 4\hat{j} + 3\hat{k}$,$\vec{b} = 8\hat{i} - 12\hat{j} - 9\hat{k}$,and $\vec{c} = 33\hat{i} - 4\hat{j} - 24\hat{k}$ represent the coterminous edges of a parallelepiped,then its volume 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 $\vec{c} = (0, 1, 0)$:

Let $\alpha \in \mathbb{R}$ and the three vectors $\vec{a} = \alpha \hat{i} + \hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + \hat{j} - \alpha \hat{k}$,and $\vec{c} = \alpha \hat{i} - 2\hat{j} + 3\hat{k}$. Then the set $S = \{ \alpha : \vec{a}, \vec{b}, \text{ and } \vec{c} \text{ are coplanar} \}$

The volume of the parallelepiped formed by the vectors $\hat{i} + m \hat{j} + \hat{k}$,$\hat{j} + m \hat{k}$,and $m \hat{i} + \hat{k}$ becomes minimum when $m$ is

If $\vec{\alpha}$ is a unit vector, $\vec{\beta}=\hat{i}+\hat{j}-\hat{k}$, and $\vec{\gamma}=\hat{i}+\hat{k}$, then the maximum value of $[\vec{\alpha} \vec{\beta} \vec{\gamma}]$ is