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)

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
\(
\begin{aligned}
& \left|\begin{array}{lll}
x-0 & y-0 & z-0 \\
0-0 & b-0 & c-0 \\
a-0 & 0-0 & c-0
\end{array}\right|=0 \Rightarrow\left|\begin{array}{lll}
x & y & z \\
0 & b & c \\
a & 0 & c
\end{array}\right|=0 \end{aligned}\)
\(\therefore x(b c)-y(-a c)+z(-a b)=0 \Rightarrow b c x+a c y~-\) \(a b z=0\)
Dividing both sides by abc, we get \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}-\frac{\mathrm{z}}{\mathrm{c}}=0\)