Let $a$ and $b$ be non- zero real numbers.
Therefore, the given equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0$ implies either
$$
\begin{array}{rc}
& x^2-5 x y+6 y^2=0 \\
\Rightarrow & (x-2 y)(x-3 y)=0 \\
\Rightarrow & x=2 y \text { and } x=3 y
\end{array}
$$
represent two straight lines passing through origin.
Or $\quad a x^2+b y^2+c=0$
When $c=0$ and $a$ and $b$ are of same signs, then
$$
\begin{aligned}
& a x^2+b y^2+c=0 \\
\Rightarrow & x=0 \text { and } y=0
\end{aligned}
$$
which is a point specified as the origin.
When $a=b$ and $c$ is of sign opposite to that of $a, a x^2+b y^2+c=0$ represents a circle.
Hence, the given equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0$
may represent two straight lines and a circle.