Let's call the given line $L$, and let's take a line $M$ that passes through the origin and the point $(2t+3,1-t,5t)$ on $L$. We're going to adjust the value of $t$ to make $M$ perpendicular to $L$.
The direction vector of $L$ is $(2,-1,5)$, and the direction vector of $M$ is $(2t+3,1-t,5t)$. So, in order for $M$ to be perpendicular to $L$, we need
$$
(2t+3,1-t,5t) \cdot (2,-1,5) = 0
$$
This gives $30t+5=0$, so $t=-\tfrac16$. Putting $t=-\tfrac16$ in the equation of $L$, we get the point $P=\tfrac16(16,7,-5)$. The desired line $M$ passes through the origin and $P$, so its equation is $M(s) = \tfrac{s}{6}(16,7,-5)$.
You can confirm that $L$ and $M$ are perpendicular, because
$$
(2,-1,5) \cdot (16,7,-5) = 0
$$
If $\mathbf{n}$ is the normal vector to the given plane and $\mathbf{p}$ is the point through which the line is supposed to pass, then the equation of the line will be of the form $\mathbf{r}=\mathbf{p}+t\mathbf{n}$. You already have both those vectors so you don't need anything else.
Best Answer
Here's a sketch.
The best way to approach this is by thinking about what fundamentally defines a line: a vector and a point. In this case, we have the point, but instead of a vector in the direction of the line, we have two vectors perpendicular to the line. However, if we take the cross product of these two vectors, it will by definition be perpendicular to both of them and thus be in the direction of the line itself. Taking the cross product gives $\langle -14, 15, 2\rangle$. Then, the line can be parametrized using standard techniques as
$$\langle 3,-8,-8\rangle+t\cdot \langle-14, 15, 2\rangle=\langle 3-14t, -8+15t, -8+2t\rangle.$$