Determine an equation of the line of intersection of the planes 4x − 3y − z = 1 and 2x + 4y + z = 5.

Answer 1

Add the 2 equations ...
6x + y = 6
y = 6-6x

let x = 0 , y = 6
2(0) + 4(6) + z = 5 --> z = -19)
we have a point (0,6,-19) on the line of intersection

let x = 1, y = 0
2(1) + 4(0) + z = 5 --> z = 3
and (1,0,3) is another point on that line

So a direction vector of that line is (1, -6, 22)
and using the point (1,0,3)

x = 1 + t
y = -6t
z = 3 + 22t

or, in symmetric form:

(x-1)/1 = y/-6 = (z-3)/22