Line of Intersection of 2 Planes:

P1: 5x – 3y – 2z = 7, and P2: 2x – y + 3z = 4.

The line of intersection, L, lies in both planes, and hence is perpendicular to the normal vectors of both planes. The cross product, n₁ ´ n₂, is also perpendicular to n₁ and n₂, so it can be used as a reference vector in the direction of the line.

From the equations, we can determine the normal vectors:

n₁ = 5i – 3j – 2k, and

n₂ = 2i – j + 3k.

Now we form the cross product:

= –11i – 19j + k

Now we need a point A on the line. Let's set x to 0 in each equation and solve for y_A and z_A.

From P1: 0 – 3y_A –2z_A = 7

From P2: 0 – y_A + 3z_A = 4

Solving, we get

, and .

We can use this point in setting up the equations of the line:

x = 0 –11t, y = –29/11 – 19t, z = 5/11 + t

or in symmetric form,