If from a point $P(a,b,c)$, perpendiculars $PA$ and $PB$ are drawn to the $yz$ and $zx$ planes respectively, then the equation of the plane $OAB$ (where $O$ is the origin) is:

1. Coordinate of A: $P(a,b,c)$ projected on $yz$ -plane ($x=0$) gives $A(0,b,c)$.2. Coordinate of B: $P(a,b,c)$ projected on $zx$ -plane ($y=0$) gives $B(a,0,c)$.3. Coordinate of O: Origin $(0,0,0)$.The equation of a plane passing through $(0,0,0)$ is $Ax+By+Cz=0$.Since it passes through $A(0,b,c)$: $B(b)+C(c)=0⟹B=-Cc/b$.Since it passes through $B(a,0,c)$: $A(a)+C(c)=0⟹A=-Cc/a$.Substituting $A$ and $B$:
$$-\frac{Cc}{a}x-\frac{Cc}{b}y+Cz=0$$
Divide by $C$ (assuming $C\ne 0$):
$$-\frac{c}{a}x-\frac{c}{b}y+z=0$$
Multiply by $-ab$:
$$bcx+cay-abz=0$$