Find rectangular Cartesian coordinate system Find rectangular Cartesian coordinate system in $\mathbb{R}^3$  to bring quadratic surface $2y^2-3z^2+4xz-12y+15=0 $ to standard form.
After splittig off square the equation can be rewritten as follows $2(y-3)^2-3z^2+4xz=3$.
What to do with mixed term  $4xz$?  
 A: Just because it's kind of fun, here's a method using the Matrix Representations of Conic Sections. Essentially, for 
$$ Q(x,z) = -3z^2 + 4xz$$
We can write 
\begin{align}
Q(x,z) &= \begin{pmatrix}
  x & z
\end{pmatrix}
\begin{pmatrix}
  0 & 2\\
  2 & -3
\end{pmatrix}
\begin{pmatrix}
  x \\ z
\end{pmatrix}
\end{align}
We can obtain the linear transformation that eliminates the cross-term $4xz$ by diagonalizing the above matrix (or in general, the best we can do is the Jordan Normal Form).
In general, the normal form for a matrix $M$ looks like 
$$ M = S J S^{-1}$$
where $J$ is a block-diagonal matrix (in this case, hopefully a completely diagonal matrix). 
Wolfram Alpha leads to 
\begin{align}
\underbrace{\begin{pmatrix}
  0 & 2\\
  2 & -3
\end{pmatrix}}_{M} &= 
\underbrace{
\begin{pmatrix}
  -1 & 2\\
  2 & 1
\end{pmatrix}}_{S}
\underbrace{
\begin{pmatrix}
  -4 & 0\\
  0 & 1
\end{pmatrix}}_{J}
\underbrace{
\begin{pmatrix}
  -1 & 2\\
  2 & 1
\end{pmatrix}^{-1}}_{S^{-1}}\\
&= 
\begin{pmatrix}
  -1 & 2\\
  2 & 1
\end{pmatrix}
\begin{pmatrix}
  -4 & 0\\
  0 & 1
\end{pmatrix}
\begin{pmatrix}
  -1/5 & 2/5 \\
  2/5 & 1/5
\end{pmatrix}
\end{align}
If we use the coordinate transformation 
\begin{align}
\begin{pmatrix}
  x' \\ z'
\end{pmatrix} &= S^{-1} \begin{pmatrix}
  x \\ z
\end{pmatrix}\\
&= \begin{pmatrix}
  -1/5 & 2/5 \\
  2/5 & 1/5
\end{pmatrix}
\begin{pmatrix}
  x \\ z
\end{pmatrix}\\
&= \begin{pmatrix}
  (2x - z)/5\\
  (2x + z)/5
\end{pmatrix}
\end{align}
To see this in action, notice also that 
\begin{align}
\begin{pmatrix}
  x \\ z
\end{pmatrix} &= S \begin{pmatrix}
  x' \\ z'
\end{pmatrix}\\
&= \begin{pmatrix}
  -1 & 2\\
  2 & 1
\end{pmatrix}
\begin{pmatrix}
  x' \\ z'
\end{pmatrix}\\
&= \begin{pmatrix}
  2z' - x'\\
  z' + 2x'
\end{pmatrix}
\end{align}
If we just let $y' = y - 3 \iff y = y' + 3$ as you've demonstrated, then subbing into the quadratic surface cartesian equation that you have gives, 
\begin{align}
2y^2 - 3z^2 + 4xz - 12y + 15 &= 0\\
\Rightarrow 2(y' + 3)^2 - 3(z' + 2x')^2 + 4(2z' - x')(z' + 2x') - 12(y' + 3) + 15 &= 0\\
\Rightarrow -20x'^2 + 2y'^2 + 5z'^2 - 3 &= 0
\end{align}
Where we've used Wolfram Alpha to simplify the equation for us.
EDIT: Here are some fun pictures to along with this.
If we set $y = 3$ to ignore the $y$ dependence for a moment, we essentially have rotated from this hyperbola

to this hyperbola

by using the change of coordinates given by lines 
\begin{align}
  x &= (2x - z)/5\\
  z &= (2x + z)/5
\end{align}

A: Let $A=\begin{bmatrix}
0 & 0 & 2\\
0 & 2 & 0\\
2 & 0 & -3\\
 \end{bmatrix}$ and $X=[x\;y\;z]^T$ and your equation can be written as: 
$X^TAX+[0\; -12\; \;\;\;0]X=0$. 
Since $A$ is symmetric, it can be diagnolized. You may verify that eigenvalues of $A$ are $1,2,-4$ (form a diagonal matrix $D$ with these eigenvalues on the diagonal), find the corresponding orthonormal eigenvectors (these exist because matrix is symmetric and by use of Gram Schmidt process) and form an orthogonal matrix $Q$ such that $AQ= QD\implies D=Q^TAQ$ Thus, 
$D=\begin{bmatrix}
1 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & -4\\
 \end{bmatrix}$ , $Q=\begin{bmatrix}
2/\sqrt{5} & 0 & -1/\sqrt{5}\\
0 & 1 & 0\\
1/\sqrt{5} & 0 & 2/\sqrt{5}\\
 \end{bmatrix}$ , $Q^TQ=QQ^T=I$, identity matrix.
Now let $X=QY=Q[x'\;\;y'\;\;z']^T$ so that the given equation becomes: $Y^T(Q^TAQ)Y+[0\; -12\; \;\;\;0][x'\;\;2y'\;\;-4z']^T=0$
$\implies Y^TDY-24y'=0
\implies x'^2+2y'^2-4z'^2-24y'=0 $. Thus we have removed the cross term.
 Can you take it from here?
A: Continue with $2(y-3)^2-3z^2+4xz=3$ and, to decouple the $xz$-term, determine the conic axes in the $xz$-plane with $f(x,z)= 4xz-3z^2$,
$$\frac{f_x’}{f_y’}=\frac{4z}{4x-6z}= \frac xz\implies (z+2x)(2z-x)=0$$
which shows that the axis $z= \frac12x$ of the conic is at $\theta$-angle with respect to the $x$-axis, with $\cos\theta =\frac2{\sqrt5}$ and $\sin\theta =\frac1{\sqrt5}$. Then, the equation can be written in the standard form as 
$$\left(\frac2{\sqrt5} x+ \frac1{\sqrt5}z\right)^2 + 2(y-3)^2-4\left(\frac1{\sqrt5} x-\frac2{\sqrt5}z\right)^2=3$$
