Given the parabola $C: y^{2}=4 x$ and the point $A(-1,-6)$, find the line $l$ passing through point $A$ such that $l$ intersects the parabola $C$ at points $P$ and $Q$, and the circle with diameter $PQ$ passes through the vertex of the parabola.
If the line $l$ passes through the origin, its equation is $y=6 x$. It intersects the parabola at points $P(0,0)$ and $Q\left(\frac{1}{9}, \frac{2}{3}\right)$. The circle with these two points as the diameter clearly passes through the vertex of the parabola. If the line $l$ does not pass through the origin, then the circle with $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$ as the diameter endpoints passes through the origin if and only if $O P \perp O Q$, i.e.,
$$
x_{1} x_{2}+y_{1} y_{2}=0,
$$

Given $y_{1}^{2}=4 x_{1}, y_{2}^{2}=4 x_{2}$, we have $y_{1} y_{2}=-16$.
Note that $l$ is neither perpendicular nor parallel to the $x$-axis. Assume its equation is
$$
y=k(x+1)-6(k \neq 0),
$$

and combine it with $y^{2}=4 x$. Eliminating $x$, we get
$$
k y^{2}-4 y+4(k-6)=0 .
$$

By Vieta's formulas, we have
$$
y_{1} y_{2}=\frac{4(k-6)}{k} .
$$

Thus,
$$
\frac{4(k-6)}{k}=-16 .
$$

Solving for $k$, we get $k=\frac{6}{5}$. When $k=\frac{6}{5}$,
$$
\Delta=16-16 k(k-6)>0,
$$

Therefore, the line
$$
y=\frac{6}{5}(x+1)-6
$$

or $6 x-5 y-24=0$ satisfies the given conditions.
Hence, the required lines are $6 x-y=0$ or $6 x-5 y-24=0$.