find the distance from the point A(-2,1,2) to the plane 3x-2y+5z+1 =0

The point A(-2,1,2)

The plane 3x-2y+5z+1=0

To find a distance between a point and a plane:

 The standard plane formula is Ax+By+Cz+D =0

which in our case 3x-2y+5z+1=0

The standard point (x1,y1,z1) = (-2,1,2)

Then the distance d= l Ax1+By1+Cz1+D l/ sqrt(A^2+B^2+C^2)

Then in our case d = l 3(-2)+(-2)(1)+5(2)+1 l/ sqrt(9+4+25)

==> d= l -6-2+10+1 l/ sqrt(38) = 3/sqrt(38) = 3/6.16 = 0.487