Vectors Question help please?

Question

The equations of two planes p1 and p2 are 3x + 4y = 7 and 2x - y + 2z = 1. Find
(i) the distances from the origin and from the point A(4,-3,0) to the plane p1.
(ii) the Cartesian equations of the set of points whose distances from the plane p1 are equal to the distance of the point A from the plane p1.
(iii) a vector equation of the line of intersection of the planes p1 and p2.

How did you find the points M and N on the planes p1 and p2? Please tell me

Answer 1

(iii) a vector equation of the line of intersection of the planes p1 and p2.

3x + 4y =7
2x -y + 2z =1

so x = (7 -4y)/3
then 2(7-4y)/3 -y + 2z =1
14/3 -8y/3 -y -1 =-2z
11/3 - 11y/3 = -2z
so z = 11y/6 - 11/6
then the equation of the intersection is:

x = 7/3 -4 t/3
y=t
z = 11t/6 -11/6

Answer 2

d(0) = 7/sqrt(25)= 7/5
d(A)= 10/3
(3x+4y-7)/sqrt(9+16) = 10/3
so 3x+4y-7 = 50/3 and so 9x +12y -71=0
also (3x+4y-7)/5 =-10/3 if you use distances with sign
Take two points which are on p1 and p2(intersection)
M(1,1,0) and N( 2,1/4,-11/8)
Vector OM = 1*i+1*j
Vector ON =2*i+1/4*j-11/8 *k
so Vector MN = 1*i-3/4j-11/8*k and the vector equation is
r=i+j+h( i-3/4j-11/8 k) where h is any real number which gives a point of the intersection