Intersection of Plane and Line

If you have a plane P: 3x - y + 4z = 3 and the line L: (x-1)/-2 = (y+4)/2 = (z+3)/2 and the point A(-1,0,5)
How would I be able the find the equation for the line M that passes through a and is perpendicular to P. Then to expand that to find the point of intersection of M and P.

I made a direction vector that when dotted with the direction vector of p it equals 0 so it is perpendicular. ( the vector I made is (2, -1, 2) ) so the parametric equation i get are:
x(t) = -1 + 2t
y(t) = 2t
z(t) = 5 - t
When I plug these into P to solve for t i get 8 = 0 so I am not to sure what I am doing wrong. Thank you for the help.

A plane doesn't have a direction vector, but it has a normal vector, and the normal vector of P is <3,-1,4>. So this should be the direction vector of M. Then

(x + 1) / 3 = y / (-1) = (z - 5) / 4

is the equation of M. Let a represent the quantity (x + 1) / 3 = y / (-1) = (z - 5) / 4. Then at the point of intersection we have

9a + a + 16 a = 3(x + 1) - y + 4(z - 5)

26a = 3x + 3 - y + 4z - 20

26 a = 3x - y + 4z - 17

26a = 3 - 17 = -14

a = -7/13.

So (x + 1) / 3 = y / (-1) = (z - 5) / 4 = -7/13

x = -34/13

y = 7/13

z = 37/13.