Chapter 5 - Inner Product Spaces - 5.2 Orthogonal Sets of Vectors and Orthogonal Projections - Problems - Page 361: 29

$\frac{16}{\sqrt 21}$

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 $P(-4,7,-2)$ and plan $P:x+2y-4z-2=0$ we have: $$distance=\frac{|-4.1+7.2+(-2)(-4)+(-2)|}{\sqrt 1^2+2^2+(-4)^2}=\frac{|-4+14+8-2|}{\sqrt 1+4+16}=\frac{16}{\sqrt 21}$$