Yes, you can find the equation of the line formed by the intersection of two planes in 3-space using their equations in a non-vector form as follows:

    1. Write the equations of the two planes in standard form:
       ax + by + cz = d1 (plane 1)
       ex + fy + gz = d2 (plane 2)

    2. Find the direction vector of the intersection line by taking the cross product of the normal vectors of the two planes. The direction vector will be perpendicular to both normal vectors and thus parallel to the intersection line. The direction vector can be found by:
       i. Finding the normal vectors of the two planes by taking the coefficients of x, y, and z in the plane equations, and putting them in a vector:

        n1 = (a, b, c)    (normal vector of plane 1)
 
        n2 = (e, f, g)    (normal vector of plane 2)

ii. Taking the cross product of the normal vectors:
        d = n1 x n2

3. Find a point on the intersection line by finding a solution to the system of equations formed by setting the two plane equations equal to each other:

       ax + by + cz = ex + fy + gz
       Rearranging, we get:
       ax - ex + by - fy + cz - gz = 0
       This equation represents the plane that contains the intersection line, so any point on this plane lies on the intersection line. We can find a point on this plane by choosing a value of one of the variables (e.g., x), and solving for the other two variables (y and z). This gives a 
        point in the form (x0, y0, z0) that lies on the intersection line.
    4. Write the equation of the line in parametric form using the direction vector found in step 2 and the point found in step 3:

       x = x0 + td y = y0 + ud z = z0 + vd

       where (x0, y0, z0) is a point on the intersection line found in step 3, (t, u, v) are parameters, and d is the direction vector found in step 2.
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