Question

Perpendicularlist b/w $3x+4y-z=s,6x+8y-2z=15$

Answer 1

Equation of planes : $3x+4y-z=5\&6x+8y-2z=15$
Obviously these are parallel planes as D.C's normal to both the planes are same.
Take a point (0,0,-5) as 1st plane:
Distance of a point $P(x,y,z)$ to plane
ax+by+cz+d=0 is given by
$\frac{|ax+by+cz+d|}{\sqrt{a^2+b^2+c^2}}$
Here $x_1=0,y_1=0,z_1=-5,d=-15$
$a=6,b=8c=-z$
Distante $\frac{|\left(2\right)(-5)-15|}{\sqrt{6^2+8^2+(-2{)}^2}}=\frac{5}{\sqrt{36+64+4}}=\frac{5}{\sqrt{104}}=\frac{5}{2\sqrt{26}}$