For what value of $a$ is the volume of the parallelepiped formed by the vectors $\hat{i} + a\hat{j} + \hat{k}$,$\hat{j} + a\hat{k}$,and $a\hat{i} + \hat{k}$ minimum?

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 $\vec{c} = (0, 1, 0)$:

The volume of the tetrahedron whose vertices are $A(-1, 2, 3)$, $B(3, -2, 1)$, $C(p, 1, 3)$, and $D(-1, -2, 4)$ is $\frac{16}{3}$ cubic units. Find the value of $p$.

Let $\vec{a}$ be a vector which is perpendicular to the vector $3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k}$. If $\vec{a} \times (2 \hat{i} + \hat{k}) = 2 \hat{i} - 13 \hat{j} - 4 \hat{k}$,then the projection of the vector $\vec{a}$ on the vector $2 \hat{i} + 2 \hat{j} + \hat{k}$ is

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

If $(2,3,9), (5,2,1), (1, \lambda, 8)$ and $(\lambda, 2,3)$ are coplanar,then the product of all possible values of $\lambda$ is.