Chapter 5 - Inner Product Spaces - 5.2 Orthogonal Sets of Vectors and Orthogonal Projections - Problems - Page 361: 31

$\frac{1}{\sqrt 5}$

From exercise 28, we have the formula: $$d=\frac{|x_0.a+y_0.b+z_0.c+d}{\sqrt a^2+b^2+c^2}$$ Since $P(-1,1,-1)$ and plan $P:z=2x$ we have: $$distance=\frac{|(-1).2+1.0+(-1).(-1)+0|}{\sqrt 2^2+0^2+(-1)^2}=\frac{|-2+0+1+0|}{\sqrt 4+1}=\frac{1}{\sqrt 5}$$