The line $L$ has intercepts $a$ and $b$ on the coordinate axes. The coordinate axes are rotated through a  fixed angle, keeping the origin fixed. If $p$ and $q$ are the intercepts of the line $L$ on the new axes, then $\frac{1}{a^2}-\frac{1}{p^2}+\frac{1}{b^2}-\frac{1}{q^2}$ is equal to

Equation of the line $L$ in the two coordinate systems is $\frac{x}{a}+\frac{y}{b}=1,\frac{X}{p}+\frac{Y}{q}=1$

where $(X, Y)$ are the new coordinates of a point (x, y) when the axes are rotated through a fixed angle, keeping the origin  fixed. As the length of the perpendicular from the origin has not changed.

$\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=\frac{1}{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}}}\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$

or $\frac{1}{a^2}-\frac{1}{p^2}+\frac{1}{b^2}-\frac{1}{q^2}=0$