This can be written (x^2-2x+1)+(y^2-4y+4)+(z^2-2xz+2z)=0; (x-1)^2+(y-2)^+(z^2-2xz+2z)=0.
Put z=0 and we are left with (x-1)^2+(y-2)^2=0. The only solution to this equation is when x=1 and y=2. This means that if we look along z axis towards the x-y plane we see a point at (1,2).
When z=1, we have x^2-4x+4+(y-2)^2=0=(x-2)^2+(y-2)^2, another point at (2,2).
When z=2, we have x^2-6x+9+(y-2)^2=0=(x-3)^2+(y-2)^2, point at (3,2).
When z=3, we have x^2-8x+16+(y-2)^2=0=(x-4)^2+(y-2)^2, point at (4,2). And so on.
When z=-1, we have x^2+(y-2)^2=0, point at (0,2).
In fact, the equation can be written (x-(z+1))^2+(y-2)^2=0.
From this it follows that, from the viewpoint of the z axis, the shape traces the line y=2.
Note that if y doesn't equal 2, the graph doesn't exist. Therefore, the plane y=2 for all x and z is the only place where the graph exists. When y=2, x=z+1 traces a straight line in the x-z plane. The figure cannot be a cone. If the expression in x, y and z equalled non-zero (e.g., 1) then at z=0 there would have been a circle of radius 1 centre (1,2). With increasing values of z, the centre would shift along the line x=z+1, but would keep the same radius, so the shape would be a tube. There would be no vertex. As the constant 1 shrinks to zero the diameter of the tube decreases until it becomes a line. To be a cone the constant would need to be a (quadratic) expression in z. (The standard equation for a circular cone is ((x-h)^2+(y-k)^2=a^2(z-j)^2, vertex at (h,k,j), and the radius of the circular base is a. The complete shape is two cones joined at their common vertex like a diabolo.) The expression's changing values would affect the diameter of the circle. When the expression evaluated to zero, the left-hand side expression would describe a point, the vertex of the cone. The sign of z^2 in the question needs to be negative to qualify as the equation of a cone.