Find the equation of the line that passes through A(1,-4, 2) and is parallel to the intersection line of the two planes x - 2y + 3z - 1 = 0 and x - 4y+ 2z - 8 = 0.

What I did:

First I set the first and second equations to [1] and [2]:

x - 2y + 3z - 1 = 0 [1]

x - 4y+ 2z - 8 = 0 [2]

I then multiply [1] by 2 and use elimination to get rid of the y variable for now:

2x - 4y + 6z - 2 = 0

x - 4y + 2z - 8 = 0

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x + 4z + 6 = 0 [3]

I'll then let z = t to solve for x in equation [3]:

x + 4t + 6 = 0

x = -4t - 6

Now I substitute z = t and x = -4t - 6 into equation [1] to solve for y:

-4t - 6 - 2y + 3t - 1 = 0

y = (-1/2)t - 7/2

Now that I have the values of all the unknowns, I first express it in parametric form:

x = -4t - 6

y = (-1/2)t - 7/2

z = t

Knowing this, finally, the direction vector for the line that passes through A(1, -4, 2) can be expressed:

(x,y,z) = (1, -4, 2) + t(-4, 1/2, 1)

I just wanted to know, did I do this correctly? I feel as if I did something wrong. If I did, can you point where I went wrong? Thank you in advance.