Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta =0$ The direction ratios of the line are $3,-5,2$. The direction ratios of the normal to the plane are $1,3,-\alpha$. The given line is perpendicular to the normal of plane. $\begin{array}{rl}\therefore & 3(1)+(-5)(3)+2(-\alpha )=0\\ & \Rightarrow 3-15-2\alpha =0\\ & \Rightarrow -12-2\alpha =0\\ & \Rightarrow \alpha =-6\end{array}$ Also, point $(2,1,-2)$ lies on the plane $\begin{array}{rl}& x+3y-\alpha z+\beta =0\\ & \Rightarrow 2+3-(-6)(-2)+\beta =0\\ & \Rightarrow 2+3-12+\beta =0\\ & \Rightarrow \beta =7\\ \therefore & {\alpha }^2+\alpha \beta +{\beta }^2=36-42+49=43\end{array}$