A plane $\pi$ contains the line $L_1:\frac{y}{b}+\frac{z}{c}=1,x=0$ and is parallel to the line $L_2:\frac{a}{a}-\frac{z}{c}=1,y=0$ then

Equation of $L_1=\frac{x}{0}=\frac{y}{-b}=\frac{z-c}{c}$

Equation of $L_2=\frac{x}{a}=\frac{y}{0}=\frac{z+c}{c}$

Direction of $\pi$ is $\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 0& -b& c\\ a& 0& c\end{array}\right|=-bc\widehat{i}+ac\widehat{j}+ab\widehat{k}$ 

Equation of is

$-bc(x-0)+ac(y-0)+ab(z-c)=0$

$-\frac{x}{a}+\frac{y}{b}+\frac{z}{c}-1=0$

Distance between $L_1\mathrm{\&}{L}_2$ is

$=\left|\frac{(0,0,2c),(bc,ac,-ab)}{\sqrt{b^2{c}^2+c^2{a}^2+a^2{b}^2}}\right|$

$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=64$