Line for set of three-dimensional vectors If there is a set for 3D vectors $v$ where
$ v \times \begin{pmatrix} -1 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 5 \\ -27 \\ 8 \end{pmatrix}$
is a line, what is this line's equation?
I'm not sure how to solve this, except for setting $v = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ and maybe having a linear system of equations, but I don't see how I could use that to solve the problem.

$
\newcommand{\i}{\hat{\mathbf{i}}}
\newcommand{\j}{\hat{\mathbf{j}}}
\newcommand{\k}{\hat{\mathbf{k}}}
\newcommand{\v}{\vec{\mathbf{v}}}
$
Let $v =  \v$ and 
$$
\v = \begin{bmatrix} a \\ b \\ c \end{bmatrix},
\quad 
\i = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix},
\quad 
\j = \begin{bmatrix} 0 \\ 1 \\ 0  \end{bmatrix},
\quad 
\k = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}
$$
Then
$$
v \times \begin{bmatrix} -1 \\ 1 \\ 4 \end{bmatrix} 
=
\begin{bmatrix} 5 \\ -27 \\ 8 \end{bmatrix}
$$
can be rewritten as 
$$
\v \times \begin{bmatrix} -1 \\ 1 \\ 4 \end{bmatrix} =
\begin{bmatrix} a \\ b \\ c \end{bmatrix} 
\times 
\begin{bmatrix} -1 \\ 1 \\ 4 \end{bmatrix}
=
\begin{bmatrix} 5 \\ -27 \\ 8 \end{bmatrix}
$$
Let us expand vector product:
$$
\begin{aligned}
\v \times \begin{bmatrix} -1 \\ 1 \\ 4 \end{bmatrix} 
=
\begin{vmatrix} 
\i & \j & \k \\
a & b & c \\
-1 & 1 & 4
\end{vmatrix}
=
\begin{vmatrix} b & c \\ 1 & 4 \end{vmatrix} \i-
\begin{vmatrix} a & c \\ -1 & 4 \end{vmatrix} \j+
\begin{vmatrix} a & b \\ -1 & 1 \end{vmatrix} \k
=
\begin{bmatrix} 5 \\ -27 \\ 8 \end{bmatrix}
\end{aligned}
$$
Thus, we conclude 
$$
\begin{cases}
4 b - c = 5 \\ 
4 a + c = 27 \\
a + b = 8
\end{cases}
\implies 
\begin{cases}
4 b - c = 5 \\ 
4 a + 4b = 32 \\
a  = 8 - b
\end{cases}
\implies 
\begin{cases}
4 b - c = 5 \\ 
32 + 8b = 32 \\
a  = 8 - b
\end{cases}
\implies 
\begin{cases}
c = -5 \\ 
b = 0 \\
a  = 8
\end{cases}
$$
Finally, we write 
$$
\bbox[9px, border:3px solid #FF0000]{v =   \begin{pmatrix} 8 \\ 0 \\ -5 \end{pmatrix}}
$$
