Take the cross product of the given vectors to get one that is perpendicular to both of them.
[tex]\langle 3, 5, -2 \rangle \times \langle 2, 1, 0 \rangle = \langle 2, -4, -7 \rangle[/tex]
The line parallel to this vector and passing through the origin is given by the vector function
[tex]\vec r(t) = \langle 2, -4, -7 \rangle t[/tex]
where [tex]t\in\Bbb R[/tex], and hence parametric equations
[tex]\begin{cases}x(t) = 2t \\ y(t) = -4t \\ z(t) = -7t \end{cases}[/tex]
Translate this line by the vector [tex]\langle1,0,1\rangle[/tex] to make it pass through the given point. So the line we want has vector function
[tex]\vec r(t) = \langle 2, -4, -7 \rangle t + \langle 1, 0, 1 \rangle[/tex]
and parametric equations
[tex]\boxed{\begin{cases} x(t) = 2t + 1 \\ y(t) = -4t \\ z(t) = -7t + 1 \end{cases}}[/tex]
