If d is the distance between two parallel planes $2\mathrm{x}+\mathrm{y}+2\mathrm{z}=8\text{ and }4\mathrm{x}+2\mathrm{y}+4\mathrm{z}+5=0$,then the value of 2d + 3 is

We know that the distance between parallel planes
$\mathrm{ax}+\mathrm{by}+\mathrm{cz}+{\mathrm{d}}_1=0 \mathrm{and} \mathrm{ax}+\mathrm{by}+\mathrm{cz}+{\mathrm{d}}_2=0 \mathrm{is} \mathrm{given} \mathrm{by} \frac{\left|{\mathrm{d}}_1-{\mathrm{d}}_2\right|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}$
Here, the planes are$2\mathrm{x}+\mathrm{y}+2z-8=0\text{ and }2\mathrm{x}+\mathrm{y}+2z+\frac{5}{2}=0$
$\therefore \mathrm{d}=\left|\frac{\left|-8-\frac{5}{2}\right|}{\sqrt{2^2+{\mathrm{l}}^2+2^2}}\right|=\frac{7}{2}\Rightarrow 2\mathrm{d}+3=10$