Step 1:

The given points are $O(0,0,0)$ and $A(1,2,-3)$

$\Rightarrow\:$ Direction ratio $(d.r.)$ of $OA\: is\:\:(x_2-x_1),(y_2-y_1),(z_2-z_1)$

$i.e.\:\:(1-0),(2-0),(-3-0)=(1,2,-3)$

Since $A$ is foot of perpendicular from origin on the plane,

$OA$ is also $\perp$ to the plane.

$\Rightarrow\:\overrightarrow n=(1,2,-3)$

Step 2:

We know that the equation of the plane passing through $(x_1,y_1,z_1)$ is

$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$

Where $(a,b,c)$ are the direction ratio of normal $(\overrightarrow n)$=$(1,2,-3)$ and the point $A$ is $(1,2,-3)$.

$\therefore$ The equation of the required plane is $1(x-1)+2(y-2)-3(z+3)=0$

$\Rightarrow\:x-1+2y-4-3z-9=0$

On simplifying

$x+2y-3z-14=0$

Hence equation of required plane is $x+2y-3z-14=0$