Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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 $b=-\frac{3}{2}c$.

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

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

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

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

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