The old coordinates are expressed through the new ones by…

The old coordinates are expressed through the new ones by $3x=x'+2y'+2z'+3$, $3y=2x'+y'-2z'+6$, $3z=2x'-2y'+z'-3$. Find the equation of the surface $7x^2+6y^2+5z^2$ $-\,4yz-4xy$ $-\,6x-24y+18z+30=0$ in the new coordinates and identify it.

Step 1: Structure of the transformation

\[ \vec r=P\vec r\,'+\vec r_0,\qquad P=\frac13\begin{pmatrix}1&2&2\\2&1&-2\\2&-2&1\end{pmatrix},\qquad \vec r_0=(1,2,-1). \]

The rows of $3P$ are pairwise orthogonal with length $3$, so $P$ is orthogonal and the change is a motion. The columns of $P$ are eigenvectors of the quadratic-form matrix $A=\begin{pmatrix}7&-2&0\\-2&6&-2\\0&-2&5\end{pmatrix}$ with eigenvalues $3,6,9$, so $P^TAP=\operatorname{diag}(3,6,9)$.

Step 2: Full substitution

Write the surface as $\vec r^{\,T}A\vec r+\vec l\cdot\vec r+30=0$ with $\vec l=(-6,-24,18)$. Then

\[ \vec r^{\,T}A\vec r=\vec r\,'^{T}(P^TAP)\vec r\,'+2\,(A\vec r_0)\cdot(P\vec r\,')+\vec r_0^{\,T}A\vec r_0 . \]

Here $A\vec r_0=(3,12,-9)$, so $2A\vec r_0=(6,24,-18)=-\vec l$. The linear terms therefore cancel: $O'$ is the centre. The constant becomes

\[ \vec r_0^{\,T}A\vec r_0+\vec l\cdot\vec r_0+30=36-72+30=-6. \]

\[3x'^2+6y'^2+9z'^2=6\quad\Longleftrightarrow\quad\frac{x'^2}{2}+\frac{y'^2}{1}+\frac{z'^2}{2/3}=1\]

Step 3: Identification

This is an ellipsoid with semi-axes $a=\sqrt2$, $b=1$, $c=\sqrt{2/3}=\tfrac{\sqrt6}{3}$. Its centre is the new origin $O'$, which is $(1,2,-1)$ in old coordinates.

Check: the centre equations $\nabla F=0$, i.e. $2A\vec r+\vec l=0$, give $\vec r=(1,2,-1)$. The value of the left-hand side there is $-6$, consistent with $F'=-6$.