Consider the lines ${\mathrm{L}}_1:\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}+3}{1}, {\mathrm{L}}_2:\frac{\mathrm{x}-4}{1}=\frac{\mathrm{y}+3}{1}=\frac{\mathrm{z}+3}{2}$ and the planes ${\mathrm{P}}_1:7\mathrm{x}+\mathrm{y}+2\mathrm{z}=3,$ ${\mathrm{P}}_2:3\mathrm{x}+5\mathrm{y}-6\mathrm{z}=4.$ Let $\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}$ the equation of the plane passing through the point of intersection of lines ${\mathrm{L}}_1\text{ and }{\mathrm{L}}_2$ and perpendicular to planes ${\mathrm{P}}_1\text{ and }{\mathrm{P}}_2$. then $(\mathrm{a}+\mathrm{d})-(\mathrm{b}+\mathrm{c})=\mathrm{\_\_\_\_\_}$