Is there a more conceptual proof of this fact? Equip ${\mathbb R}^3$ with the usual scalar product $(.|.)$. Let $A$ be the matrix
$$
A=
\left(\begin{matrix}
1 & 2 & 3 \\
-2 & 4 & 5 \\
-3 & -5 & 6 \\
\end{matrix}\right)
$$
With the help of a computer, I found an ugly, wholly computational proof of the following
fact :
Theorem There is no orthogonal basis $(b_1,b_2,b_3)$ such that
$(Ab_1|b_2)=(Ab_2|b_3)=(Ab_3|b_1)=0$.
(I argue as follows : there must be at least one index $i$ such that
$Ab_i$ is not a multiple of $b_i$, say $i=1$. Then there is a constant
$c$ such that $b_2=c(b_1 \wedge Ab_1)$, and we assume without loss
that $c=1$. Similarly we may assume $b_3=b_1 \wedge b_2$. We may express
everything in terms of the coordinates $x,y,z$ of $b_1$. Finally
we have a system of two polynomial equations in $x,y,z$ (coming
from $(Ab_2|b_3)=(Ab_3|b_1)=0$ ), which can be shown to have no real
solution by a formal computing system). 
Does anyone know a more interesting and conceptual method ?

You are essentially asking if there exists an orthogonal matrix $B$ such that
$$
B^TAB=\pmatrix{x&2a&0\\ 0&y&2b\\ 2c&0&z}\tag{1}
$$
for some $x,y,z,a,b,c\in\mathbb R$. By splitting into symmetric and skew-symmetric parts, we can rewrite the equation as
\begin{align*}
B^T\pmatrix{1\\ &4\\ &&6}B &= \pmatrix{x&a&c\\ a&y&b\\ c&b&z},\tag{2}\\
B^T\pmatrix{0&2&3\\ -2&0&5\\ -3&-5&0}B &= \pmatrix{0&a&-c\\ -a&0&b\\ c&-b&0}.\tag{3}
\end{align*}
Take the Frobenius norm on both sides of $(2)$ and $(3)$, we get
\begin{align*}
1^2+4^2+6^2 &= x^2+y^2+z^2+2(a^2+b^2+c^2),\\
2(2^2+3^2+5^2) &= 2(a^2+b^2+c^2).
\end{align*}
Therefore $x^2+y^2+z^2=-23$, which is insolvable. Hence the orthogonal basis in question does not exist.
