Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Any plane passing through $(-2,2,2)$ is $A(x+2)+B(y-2)+C(z-2)=0$ $\because$ It passes through $(2,-2,-2)$ $\Rightarrow 4A-4B-4C=0$ It is parallel to $9x-13y-3z=0$ $\begin{array}{lll}\therefore & 9A-13B-3C=0& \dots \text{ (ii) }\end{array}$ Solving Eqs. (i) and (ii), $$\begin{array}{rl}& \frac{A}{12-52}=\frac{B}{-36+12}=\frac{C}{-52+36}\\ \Rightarrow \frac{A}{-40}=\frac{B}{-24}=\frac{C}{-16}& \end{array}$$ $\therefore$ Required equation of plane is $-40(x+2)-24(y-2)-16(z-2)=0$ $\begin{array}{lr}\Rightarrow & -40x-80-24y+48-16z+32=0\\ \Rightarrow & 40x+24y+16z=0\\ \Rightarrow & 5x+3y+2z=0\end{array}$