Write the vector equation of the plane 3x-y-7z=5. Find the equations of the planes through the intersection of the planes x+3y+6=0 and 3x-y-4z=0,whose perpendicular distance from the origin is equal to 1.

Vector equation is :

r.(3i - j - 7k) = 5

Let the plane be

(x+ 3y + 6) + a(3x - y - 4z) = 0    ----------------- 1

This plane is at a distance of 1 from (0,0,0)

so 6/root((1+3a)² + (3-a)² + (4a)²)  =  1

On solving we get

 a = 1 , -1

Substitute in eq 1 to get the plane