If A and B are the feet of the perpendiculars | vedclass

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(A) $1$. The point $P$ is $(1, 2, 3)$.$2$. The foot of the perpendicular from $P(1, 2, 3)$ to the $YZ$-plane $(x=0)$ is $A(0, 2, 3)$.$3$. The foot of

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

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(A) $1$. The point $P$ is $(1, 2, 3)$.$2$. The foot of the perpendicular from $P(1, 2, 3)$ to the $YZ$-plane $(x=0)$ is $A(0, 2, 3)$.$3$. The foot of the perpendicular from $P(1, 2, 3)$ to the $ZX$-plane $(y=0)$ is $B(1, 0, 3)$.$4$. The plane passes through the origin $O(0, 0, 0)$, $A(0, 2, 3)$, and $B(1, 0, 3)$.$5$. The equation of a plane passing through the origin is $ax + by + cz = 0$.$6$. Substituting $A(0, 2, 3)$: $2b + 3c = 0 \implies b = -\frac{3}{2}c$.$7$. Substituting $B(1, 0, 3)$: $a + 3c = 0 \implies a = -3c$.$8$. Let $c = -2$, then $a = 6$ and $b = 3$.$9$. The equation is $6x + 3y - 2z = 0$.

If $A$ and $B$ are the feet of the perpendiculars drawn from $(1, 2, 3)$ to planes $YZ$ and $ZX$ respectively, then the equation of the plane passing through the points $A$, $B$ and the origin $(0, 0, 0)$ is

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