If A and B are the feet of the perpendiculars drawn from $(1, 2, 3)$ to planes $YZ$ and $ZX$, then the equation of the plane passing through the points A, B and the origin is

Let the given point be $P(1, 2, 3)$.

The foot of the perpendicular from $P$ to the $YZ$-plane ($x = 0$) is $A(0, 2, 3)$.

The foot of the perpendicular from $P$ to the $ZX$-plane ($y = 0$) is $B(1, 0, 3)$.

The equation of the plane passing through the origin $(0, 0, 0)$ is $ax + by + cz = 0$.

Since the plane passes through $A(0, 2, 3)$ and $B(1, 0, 3)$, we have:

$2b + 3c = 0$

$a + 3c = 0$

Solving these equations, we get $a = -3c$ and $b = -\dfrac{3}{2}c$.

Substituting the values of $a$ and $b$ in the equation of the plane:

$-3cx - \dfrac{3}{2}cy + cz = 0$

Dividing by $-\dfrac{c}{2}$, we get the required equation of the plane:

$6x + 3y - 2z = 0$

Answer: $6x + 3y - 2z = 0$