Chapter 5 - Inner Product Spaces - 5.2 Orthogonal Sets of Vectors and Orthogonal Projections - Problems - Page 361: 30

$\frac{7}{\sqrt 11}$

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 $P(0,-1,3)$ and plan $P:3x-y-z=5$ we have: $$distance=\frac{|0.3+(-1).(-1)+3(-1)+(-5)|}{\sqrt 3^2+(-1)^2+(-1)^2}=\frac{|0+1-3-5|}{\sqrt 9+1+1}=\frac{7}{\sqrt 11}$$