Conical Equation A quadric surface has the following equation: $2 x^2+3 y^2+3 z^2+J+16 x−18 y−6 z=0$ 
Enter a value of $J$ for which the quadric is :
a) A single point
b) The empty set
Working - 
I honestly have no clue on how to answer either of the two questions. All that I do know is that I can complete the square to get $2(x+4)^2+3(y-3)^2+3(z-1)^2+J=0$ but do not know how to find a value of J such that the equation equals both a single point or the empty set. I do however also know that the value for J to make the equation a single point is a number however. Any suggestions are appreciated. Please edit my question for any clarification. Thank you 

$$2(x+4)^2+3(y-3)^2+3(z-1)^2+J=0$$
is not correct.
We have
$$\begin{align}&2x^2+3y^2+3z^2+J+16x-18y-6z=0\\&\iff 2x^2+16x+3y^2-18y+3z^2-6z+J=0\\&\iff 2x^2+16x+32+3y^2-18y+27+3z^2-6z+3=32+27+3-J\\&\iff 2(x+4)^2+3(y-3)^2+3(z-1)^2=62-J\end{align}$$
