Let the equation of the requierd plane be

$a x+by+cz+d=0$

It is given that the plane is equally inclined coordinate axes.

Hence its direction cosines are $(1,1,1)$

=> the equation of the plane is $(x+y+z+d)=0$

It is also given that $\perp$ distance from the origin is $3 \sqrt 3$

$(ie)\;\bigg| \large\frac{d}{a^2+b^2+c^2} \bigg| $ $= 3\sqrt 3$

Substituting for $(a,b,c)$ as $(1,1,1)$

$\bigg| \large\frac{d}{1^2+1^2+1^2} \bigg| $ $= 3\sqrt 3$

$\large\frac{d}{\sqrt 3}$$=\pm 3 \sqrt 3$

$=>d=\pm 9$

Therefore the equation of the plane is $x+y+z=\pm 9$