The angle between the planes represented by $2{\mathrm{x}}^2-6{\mathrm{y}}^2-12{\mathrm{z}}^2+18\mathrm{yz}+2\mathrm{zx}+\mathrm{xy}=0\text{ is }{\mathrm{Cos}}^{-1}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)$ then least value of m + n is______

The equation $2{\mathrm{x}}^2-6{\mathrm{y}}^2-12{\mathrm{z}}^2+18\mathrm{yz}+2\mathrm{zx}+\mathrm{xy}=0$ represents two planes and they are$$\begin{gathered} 2{\mathrm{x}}^2-6{\mathrm{y}}^2+\mathrm{xy}-12{\mathrm{z}}^2+18\mathrm{yz}+2\mathrm{zx}=0 \\ \Rightarrow \left(x+2y+mz\right)\left(2x-3y+nz\right)=0 \\ \Rightarrow 2m+n=2,-3m+2n=18 \\ \therefore m=-2,n=6 \end{gathered}$$

$2\mathrm{x}-3\mathrm{y}+6\mathrm{z}=0\dots .\left(1\right); \mathrm{x}+2\mathrm{y}-2\mathrm{z}=0...\left(2\right)$

 If $\mathrm{\theta }$ is the angle b/w the planes (1) and (2) then

$\begin{array}{l}\cos \mathrm{\theta }=\frac{\left|\right(2\left)\right(1)+(-3\left)\right(2)+6(-2\left)\right|}{\sqrt{2^2+(-3{)}^2+6^2}\sqrt{1^2+2^2+2^2}}=\frac{16}{21}\\ \therefore \mathrm{\theta }={\mathrm{Cos}}^{-1}\frac{16}{21}\end{array}$