Vector Equation of a Line of a line perpendicular to a plane

State the vector equation of a line that that is perpendicular to the $yz$ plane and passes through $Q(-6, 5, 0)$.

Obviously if the line is perpendicular to the plane it would have the same direction vector as that of the normal of the plane. But I do not know how to figure this out.... Please help!

As it is perpendicular to the $yz$ plane, its direction vector is $(1,0,0)$, so its parametric equation is $$(x,y,z)=(-6,5,0)+t(1, 0, 0)=(t-6,5,0).$$
If you know two vectors which lie on a plane, then you can use the cross product to find the normal of the plane. In your question, we want to find the normal to the $yz$-plane. Two vectors which lie on the $yz$-plane are $e_2=(0,1,0)$ and $e_3=(0,0,1)$. Since $e_2\times e_3=e_1=(1,0,0)$, the normal to the plane is $(1,0,0)$.
To find the vector equation of any line, you need a point on the line, and a direction vector of the line: $$l(t)=(-6,5,0)+t(1,0,0).$$
You could also have come up with $e_1$ as the normal to the $yz$-plane by realizing that the $x, y$ and $z$ axis are perpendicular, so the $x$-axis (which is spanned by the vector $e_1$) must be perpendicular to the $yz$-plane.