If $a, b, c$ are three non-coplanar vectors and $d$ is any unit vector, then $|(a \cdot d)(b \times c) + (b \cdot d)(c \times a) + (c \cdot d)(a \times b)| = $

$a, b, c$ are three non-zero,non-coplanar vectors and $p, q, r$ are three other vectors such that $p = \frac{b \times c}{a \cdot (b \times c)}$,$q = \frac{c \times a}{a \cdot (b \times c)}$,$r = \frac{a \times b}{a \cdot (b \times c)}$. Then $[p, q, r]$ equals

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 parallelepiped whose conterminous edges are $\vec{a} = \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = 2\hat{i} - 4\hat{j} + 5\hat{k}$,and $\vec{c} = 3\hat{i} - 5\hat{j} + 2\hat{k}$ is:

Consider $\overrightarrow{r}, \overrightarrow{a}, \overrightarrow{b}$ and $\overrightarrow{c}$ are non-zero vectors such that $\overrightarrow{r} \cdot \overrightarrow{a}=0$,$|\overrightarrow{r} \times \overrightarrow{b}|=|\overrightarrow{r}||\overrightarrow{b}|$,and $|\overrightarrow{r} \times \overrightarrow{c}|=|\overrightarrow{r}||\overrightarrow{c}|$. Then,the scalar triple product $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$ is:

If the vectors $2i - j + k$,$i + 2j - 3k$,and $3i + aj + 5k$ are coplanar,find the value of $a$.