Question

Write the equation of a plane which is at a distance of 5√3 units from origin and the normal to which is equally inclined to coordinate axes.

Answer 1

check the attachment

Answer 2

Given: distance from origin: 5$\sqrt{3\\}$ units

to find: equation of plane

Let a, b, c be direction cosine of the line

a²+b²+c²=1

ax+by+cz+d=0

now, since a= b= c (equally inclined),

so a= b= c= 1/√3

so,

x/√3 + y/√3 +z/√3 ±5√3=0

So we get two equations, those are

x +y +z  +15 =0,

x +y +z  -15 =0

hence,

       mod(d)/√a²+b²+c²  = 5√3

       ∴d = ±5√3