Find the plane which touches the cone $x^2+2y^2-3z^2+2yz-5zx+3xy=0$ along the generator whose direction ratios are $1,1,1.$
Let the plane touches the cone at $(\alpha,\beta,\gamma)$.
We know that the equation of the tangent plane to the cone $ax^2+by^2+cz^2+2fyz+2gzx+2hxy=0$ at the point $(\alpha,\beta,\gamma)$ is
$x(a\alpha+h\beta+g\gamma)+y(h\alpha+b\beta+f\gamma)+z(g\alpha+f\beta+c\gamma)=0$
The equation of the cone is $x^2+2y^2-3z^2+2yz-5zx+3xy=0$.Here $a=1,b=2,c=-3,f=1,g=-\frac{5}{2},h=\frac{3}{2}.$
The equation of the tangent plane is $x(\alpha+\frac{3}{2}\beta-\frac{5}{2}\gamma)+y(\frac{3}{2}\alpha+2\beta+\gamma)+z(-\frac{5}{2}\alpha+\beta-3\gamma)=0$
The given plane is parallel to the generator whose direction ratios are $1,1,1$ or whose direction cosines are $\frac{1}{\sqrt3},\frac{1}{\sqrt3},\frac{1}{\sqrt3}$
So $\frac{1}{\sqrt3}(\alpha+\frac{3}{2}\beta-\frac{5}{2}\gamma)+\frac{1}{\sqrt3}(\frac{3}{2}\alpha+2\beta+\gamma)+\frac{1}{\sqrt3}(-\frac{5}{2}\alpha+\beta-3\gamma)=0$
$\frac{9\beta}{2}-\frac{9\gamma}{2}=0$
$\beta=\gamma$
I am stuck here.Please help me.