Show that all plane sections of the conicoid $ax²+by²+cz²= 1$ which are rectangular hyperbolas and which pass through the point $(\alpha,\beta,\gamma)$ touch the cone $\frac{(x-\alpha)^2}{b+c}$+$\frac{(y-\beta)^2}{c+a}$+$\frac{(z-\gamma)^2}{a+b}$=0
My attempt
Given equation of comicoid is $ax²+by²+cz²= 1$ Let $lx+my+nz=0$ be a plane which cuts the conicoid in rectangular hyperbolas. Therefore,
$(b+c)l² + (c+a)m² + (a+b)n² = 0$ ..........(1)
Now the plane passes through the point $(\alpha,\beta,\gamma)$. Therefore l$\alpha$+m$\beta$+n$\gamma$=0
Hence we can write the equation of plane as $lx+my+nz =$ l$\alpha$+m$\beta$+n$\gamma$
or, l(x-$\alpha$)+m(y-$\beta$)+n(z-$\gamma$)=0 ....... (2) I don't know how to proceed further please help me