Translate and rotate the coordinate axes to put the quadric in standard position, $2xy+2xz+2yz-6x-6y-4z=-9$ 
Referring to the image attached. Is my working correct?
The answer gives the coefficient on the other side of the equation is -1. Please show where went wrong.
Thanks.
 A: Let $A =
\begin{pmatrix}
0 & 1 & 1\\
1 & 0 & 1\\
1 & 1 & 0
\end{pmatrix} \, .
$
To write the quadric in standard form, we must write it with respect to an orthonormal set of eigenvectors, which amounts to orthogonally diagonalizing $A$.
As mentioned in the comments, you need to apply Gram-Schmidt orthogonalization to obtain an orthonormal basis for the eigenspace of the eigenvalue $-1$.
Diagonalizing $A$, we find that $P^{-1} A P = D$ where
$$
D =
\left(\begin{array}{rrr}
2 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{array}\right)
\qquad
P =
\left(\begin{array}{rrr}
1 & 1 & 0 \\
1 & 0 & 1 \\
1 & -1 & -1
\end{array}\right) \, .
$$
Let $w_i$ be the $i^\text{th}$ column of $P$.  Note that the last two columns $w_2, w_3$ are not orthogonal.  Applying Gram-Schmidt, we set $v_2 = w_2$ and compute
$$
v_3 = w_3 - \operatorname{proj}{v_2}(w_3) = w_3 - \frac{w_3 \cdot v_2}{v_2 \cdot v_2} v_2 = w_3 - \frac{1}{2} v_2 = \begin{pmatrix} -1/2\\ 1\\ -1/2 \end{pmatrix} \, .
$$
Normalizing the vectors, we obtain the orthogonal matrix
$$
Q = 
\left(\begin{array}{rrr}
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}}\\
\frac{1}{\sqrt{3}} & 0 & \frac{\sqrt{2}}{\sqrt{3}}\\
\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}}
\end{array}\right) \, .
$$
To deal with the degree $1$ terms, we compute
$$
\left(-6,\,-6,\,-4\right)
\left(\begin{array}{rrr}
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}}\\
\frac{1}{\sqrt{3}} & 0 & \frac{\sqrt{2}}{\sqrt{3}}\\
\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}}
\end{array}\right)
= \left(-\frac{16}{\sqrt{3}}, -\sqrt{2},\,-\frac{\sqrt{2}}{\sqrt{3}}\right) \, .
$$
Thus the transformed quadric is $0 = 2 x^2 - y^2 - z^2 - \frac{16}{\sqrt{3}} x - \sqrt{2}y -\frac{\sqrt{2}}{\sqrt{3}} z + 9$ and completing the square yields
\begin{align*}
0 &= 2 x^2 - y^2 - z^2 - \frac{16}{\sqrt{3}} x - \sqrt{2}y -\frac{\sqrt{2}}{\sqrt{3}} z + 9\\
&= 2\left(\left(x - \frac{4}{\sqrt{3}} \right)^2 - \frac{16}{3} \right) - \left(\left(y + \frac{\sqrt{2}}{2} \right)^2 - \frac{1}{2} \right) - \left(\left(z + \frac{1}{\sqrt{6}} \right)^2 - \frac{1}{6} \right) + 9\\
&= 2 u^2 - v^2 - w^2 - \frac{32}{3} + \frac{1}{2} + \frac{1}{6} + 9 = 2 u^2 - v^2 - w^2 - 1 \, .
\end{align*}
