At what point on the paraboloid y = x2 z2 is the tangent plane parallel to the plane 3x 2y 5z = 8?

To find the tangent plane to the surface f(x,y,z)=0 at a point (X,Y,Z) we use the following method: 

Calculate grad f = (f_x, f_y, f_z). The normal vector to the surface at the point (X,Y,Z) is grad f(X,Y,Z). The equation of a plane with normal vector n which passes through the point p is (r-p).n=0, where r=(x,y,z) is the position vector. So the equation of the tangent plane to the surface through the point (X,Y,Z) is ((x,y,z)-(X,Y,Z)).grad f(X,Y,Z)=0. 

Now in your case we have f(x,y,z)=y-x^2-z^2, so grad f=(-2x,1,-2z), and the equation of the tangent plane at the point (X,Y,Z) is 

((x,y,z)-(X,Y,Z)).(-2X,1,-2Z)=0, 

that is 

-2X(x-X)+1(y-Y)-2Z(z-Z)=0, 

i.e. 

-2Xx+y-2Zz = -2X^2+Y-2Z^2. (1) 

Now compare this equation with the plane 

x + 2y + 3z = 1. (2) 

The two planes a_1x+b_1y+c_1z=d_1, a_2x+b_2y+c_2z=d_2 are parallel when (a_1,b_1,c_1) is a multiple of (a_2,b_2,c_2). So the two planes (1),(2) are parallel when (-2X,1,-2Z) is a multiple of (1,2,3), and we have 

(-2X,1,-2Z)=1/2(1,2,3) 

for X=-1/4 and Z=-3/4. On the paraboloid the corresponding y coordinate is Y=X^2+Z^2=1^4+9^4=5/2. 

So the tangent plane to the given paraboloid at the point (-1/4,5/2,-3/4) is parallel to the given plane.

Letter a, because when there are unlike charges they will always attract to each other.