Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Angle between given planes is
Cos$\theta =\left|\frac{{\stackrel{-}{n}}_1 \cdot {\stackrel{-}{n}}_2}{\left|{\stackrel{-}{n}}_1\right| \left|{\stackrel{-}{n}}_2\right|} \right|\Rightarrow \frac{1}{2}=\left|\frac{2m+m-2}{\sqrt{m^2 + 5} \sqrt{m^2 + 5}}\right|$
$\frac{1}{2}=\left|\frac{3m-2}{m^2+5}\right|\Rightarrow m^2+5=\pm \left(6m-4\right)$
$\Rightarrow m^2+5=6m-4,m^2+5=-6m+4$
$m^2-6m+9=0,m^2+6m+1=0$
${\left(m-3\right)}^2=0$
$m=3$