If $A$ and $B$ are the foot of the perpendicular drawn from the point $\mathrm{Q}(\mathrm{a}, \mathrm{b},…
If $A$ and $B$ are the foot of the perpendicular drawn from the point $\mathrm{Q}(\mathrm{a}, \mathrm{b}, \mathrm{c})$ to the planes $\mathrm{YZ}$ and $\mathrm{ZX}$ respectively, then the equation of the plane through the points $\mathrm{A}, \mathrm{B}$, and $\mathrm{O}$ is (where $\mathrm{O}$ is the origin)
$\frac{x}{a}-\frac{y}{b}-\frac{z}{c}=0$
$\frac{x}{a}+\frac{y}{b}-\frac{z}{c}=0$
$\frac{x}{a}-\frac{y}{b}+\frac{z}{c}=0$
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0$
Solution
From given data, we write $\mathrm{A} \equiv(0, \mathrm{~b}, \mathrm{c})$ and $\mathrm{B} \equiv(\mathrm{a}, 0, \mathrm{c})$ Equation of plane passing through $\mathrm{A}, \mathrm{B}$ and $\mathrm{O}$ is
$\begin{aligned}
& \left|\begin{array}{lll}
x-0 & y-0 & z-0 \\
0-0 & b-0 & c-0 \\
a-0 & 0-0 & c-0
\end{array}\right|=0 \Rightarrow\left|\begin{array}{lll}
x & y & z \\
0 & b & c \\
a & 0 & c
\end{array}\right|=0 \\
& \therefore x(b c)-y(-a c)+z(-a b)=0 \Rightarrow b c x+a c y-a b z=0
\end{aligned}$
Dividing both sides by abc, we get $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}-\frac{\mathrm{z}}{\mathrm{c}}=0$