We are given that,

A plane is at a distance of 3√3 units from the origin.

Also, the normal is equally inclined to coordinate axis.

We need to find the equation of the plane.

We know that, vector equation of a plane at a distance d from the origin is given by

lx + my + nz = d …(i)

where,

l, m and n are direction cosines of the normal of the plane.

And since, normal is equally inclined to coordinate axis, then

l = m = n

⇒ cos α = cos β = cos γ …(ii)

Also, we know that,

cos² α + cos² β + cos² γ = 1

⇒ cos² α + cos² α + cos² α = 1 [from (ii)]

⇒ 3cos² α = 1

This means,

Substituting values of l, m and n in equation (i), we get

Here, d = 3√3

So,

⇒ x + y + z = 3√3 × √3

⇒ x + y + z = 3 × 3

⇒ x + y + z = 9

Thus, the required equation of the plane is x + y + z = 9.