Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

(A) The equation of a plane passing through $(1,-1,2)$ is $a(x-1)+b(y+1)+c(z-2)=0$ If is perpendicular to the planes $2x+3y-2z=5$ and $x+2y-3z=8$. $\therefore 2\mathrm{a}+3\mathrm{b}-2\mathrm{c}=0$ and $\mathrm{a}+2\mathrm{b}-3\mathrm{c}=0$ Solving the above equation, we get, $\begin{array}{l}\frac{a}{\left|\begin{array}{cc}3& -2\\ 2& -3\end{array}\right|}=\frac{-b}{\left|\begin{array}{ll}2& -2\\ 1& -3\end{array}\right|}=\frac{c}{\left|\begin{array}{ll}2& 3\\ 1& 2\end{array}\right|}\\ \frac{a}{-5}=\frac{b}{4}=\frac{c}{1}\end{array}$ Substituting $\mathrm{a}=-5,\mathrm{b}=4$ and $\mathrm{c}=1$, we get, $-5x+4y+z=-7\Rightarrow 5x-4y-z=7$ $\therefore$ Required equation can be written as $\overline{\mathrm{r}}\cdot (5\hat{\mathrm{i}}-4\hat{\mathrm{j}}-\hat{\mathrm{k}})=7$