Find the equation of the straight line that contains a point, is perpendicular to a line , and is parallel to a plane. Find the equation of the straight line that contains the point $(3,7,5)$, is perpendicular to line $x-1=\frac{y+7}{-2}=\frac{z-2}{3}$, and is parallel to plane $2x+y-z-2=0$.
To make this possible, should the plane must be either perpendicular also or parallel to the given line?
I am trying to solve it by illustrating it first using graphing software. It seems that the plane and the given line are neither perpendicular nor parallel.
If the said condition is not needed, how can I solve the problem?
Note: This is not an exam or homework. I am reviewing for an upcoming test.
Thank you in advance for those who will answer my question.
 A: If your searched for line contains the point $(3,7,5)$, we can write it as
$$f=\begin{pmatrix}3\\7\\5\end{pmatrix}+r\begin{pmatrix}a\\b\\c\end{pmatrix}$$
With a free variable $t$ the given line can be rewritten to
$$g=\begin{pmatrix}t\\-2t-5\\3t+1\end{pmatrix}=\begin{pmatrix}0\\-5\\1\end{pmatrix}+t\begin{pmatrix}1\\-2\\3\end{pmatrix}$$
To get there, set $x=t$ and rewrite the equations to get a value for $y$ and $z$ in regard to $t$.
The given plane has the normal vector
$$ n=\begin{pmatrix}2\\1\\-1\end{pmatrix} $$
as can be seen directly from its coordinate form.
To fulfill the requirement that the line is perpendicular to the given line and parallel to the given plane we compute the cross product of the two vectors
$$\begin{pmatrix}1\\-2\\3\end{pmatrix} * \begin{pmatrix}2\\1\\-1\end{pmatrix} =\begin{pmatrix} -2\cdot (-1)-1\cdot 3\\ 3\cdot 2 - 1\cdot(-1)\\1\cdot 1- 2\cdot (-2)\end{pmatrix}=\begin{pmatrix}-1\\7\\5\end{pmatrix}$$
Thus, the final solution is
$$f=\begin{pmatrix}3\\7\\5\end{pmatrix}+r\begin{pmatrix}-1\\7\\5\end{pmatrix}$$
The only requirement here is that the cross product exist, that is that the plane is not exactly parallel to the given line.
