Find an equation for the plane that contains the two lines

l1:x=t+ 1,y= 5t– 4,z= 4t+ 3

and

l2:x= -2 –t,y= 4t– 19,z= -9 – 2t.

I rewrote this in vector form

L1 = (1,-4,3) + t(1,5,4)

L2 = (-2,-19,-9) + t(-1,4,-2)

Then i went to find the point of intersection

x = 1+s = -2-t

y = -4+5s = -19+4t

z = 3+4s = -9-2t

used x to find 's'

1+s = -2-t

s = -3-t

plugged in 's' to z

3+4(-3-t) = -9-2t

3-12-4t=-9-2t

9-4t=-9-2t

-6t=0

t = 0

plugged 't' back into x

1+s = -2-(0)

s = -3

I plugged in t into x,y and z and get the following

x = -2-(0) = -2

y = -19+4(0) = -19

z = -9-2(0) = -9

P = (-2,-19,-9)

Then i went to find the cross product

n = (1,5,4)x(-1,4,-2)

n = (-26,2,9)

Then finally went ahead to find the equation.

-26(x-(-2)) + 2(y-(-19)) + 9(z-(-9)) = 0

-26x-52 + 2y+38 + 9z+81 = 0

-26x + 2y + 9z = -67

but this is WRONG!!! what am i doing wrong? Thanks!!!