The vector equation for any line is
$$r(t) = \text{point you want it to pass through} + \text{parameter} \cdot \text{velocity vector}.$$
You want it to pass through the point $P = (-1,-5,2)$ and uses the parameter $t$, so we write
$$r(t) = (-1,-5,2) + t \cdot \text{velocity vector}.$$
As it asked to set the velocity vector as the normal vector to the plane, and that is $N = (1,-5,1)$, we get
$$r(t) = (-1,-5,2) + t (1,5,1).$$
The parameter could have been anything else. We could have chosen $2t, t/7$ or $4t-3$. What difference does it make?
In the first two cases we are changing the speed at which the point walks the line. With $2t$ it walks twice as faster, with $t/7$ it walks $1/7$ slower.
The case $(4t-3)$ changes both speed and at what time you pass through the desired point. With $(4t-3)$ you'll pass through point $P$ at the time $t=3/4$. Using the parameter $t$ ensures that at time $t=0$, so to speak, you begin at point $(-1,-5,2)$.
The method (2) is correct since the normal vector to the searched plane have to be orthogonal to the vector $P_1-P_2= (-3,1,1)$ and to the normal vector to the given plane $ (8,-2,6)$.
For the method (1) note that an orthogonal vector to $ (8,-2,6)$ has components such that: $(8,-2,6)(x,y,z)^T=0$ so you can fix only one component from this equation and the other two can be fixed imposing that the plane pass through the two given points.
Best Answer
Since $L(t) = (4,0,-1) + (3,2,4)t$, the vector $v = (3,2,4)$ is a normal vector to the plane. Then using the given point $(5,-4,2)$, the equation of the plane is
$$0 = 3(x-5) + 2(y+4) + 4(z-2) $$