Answer
$d=\frac{142}{\sqrt 430}$
Work Step by Step
We need to find plane $P$ first:
$P=\{a.v+b.w\}: a,b \in R$
Substituting:
$P=\{a.(-1,0,5)+b.(3,2,-1)\\
=\{(-a,0,5a)+(3b,2b,-b)\}\\
=\{(-a+3b,2b,5a-b)\}$
Thus:
$x=-a+3b\\
y=2b\\
z=5a-b$
From the first equation, we obtain:
$a=y-x\\
b=\frac{y}{3}$
Substituting to the third equation:
$z=5(y-x)-\frac{y}{3}\\
\rightarrow 15x-14y+3z=0$
Hence, the equation of plane $P$ formed by $v$ and $w$ is: $15x-14y+3z=0$
Using the formula from Exercise 28, the distance of a point $P(x_0,y_0,z_0)$ to the plane $P..ax+by+cz+d=0$ is given by:
$$d(P,P)=\frac{|x_0.a+y_0.b+z_0.c+d|}{\sqrt a^2+b^2+c^2} \\
=\frac{(-2).15+8.(-14)+0.3+0}{\sqrt 15^2+(-14)^2+3^2}\\
= \frac{|142|}{\sqrt 225+196+9}\\
=\frac{142}{\sqrt 430}$$
Hence, $d=\frac{142}{\sqrt 430}$
