When the coordinate axes are rotated about the origin in the positive direction through an angle $\pi /4$, if the equation $25x^2+9y^2=225$ is transformed to $\alpha x^2+\beta xy+\gamma y^2=\delta ,$then ${(\alpha +\beta +\gamma -\sqrt{\delta })}^2=\mathrm{.........}$

Let (X, Y) be the new coordinates of (x, y) and angle of rotation, $\theta =45°$

Now $x=X\cos 45°-Y\sin 45°=\frac{X-Y}{\sqrt{2}},y=X\sin 45°+Y\cos 45°=\frac{X+Y}{\sqrt{2}}$

The transformed equation of $25X^2+9Y^2=225$ is $25{\left(\frac{{\displaystyle X-Y}}{{\displaystyle \sqrt{2}}}\right)}^2+9{\left(\frac{{\displaystyle X+Y}}{{\displaystyle \sqrt{2}}}\right)}^2=225$

$\Rightarrow 25(X^2-2XY+Y^2)+9(X^2+2XY+Y^2)=450\Rightarrow 34Y^2-32XY+34Y^2=450$

$\Rightarrow 17X^2-16XY+17Y^2=225\to \left(1\right)$ 

Comparing equation (1) with $\alpha X^2+\beta XY+\gamma Y^2=\delta ,$ we have $\alpha =17,\beta =-16,\gamma =17,\delta =225$

Now ${(\alpha +\beta +\gamma -\sqrt{\delta })}^2={(17-16+17-15)}^2={\left(3\right)}^2=9$