Now

${\mathbf{U}}_{\perp \mathbf{V}}$ = $\mathbf{U}-\frac{\mathbf{U}·\mathbf{V}}{\mathbf{V}·\mathbf{V}} \mathbf{V}$ = $\left[\begin{array}{c}1\\ 2\\ -5\end{array}\right]-\left(-\frac{36}{155}\right)\left[\begin{array}{c}7\\ -9\\ 5\end{array}\right]$ = $\left[\begin{array}{c}\frac{407}{155}\\ -\frac{14}{155}\\ -\frac{119}{31}\end{array}\right]$ 

and $d\=∥\mathbf{U}-\frac{\mathbf{U}·\mathbf{V}}{\mathbf{V}·\mathbf{V}} \mathbf{V}∥$ = $\frac{1}{155}\sqrt{519870}$ = $\frac{1}{155}\sqrt{3354}\sqrt{155}$ ≐ 4.65.

Alternatively, Table 1.5.1 lists a formula for finding the distance from point P to the line through points Q and R. This formula is not immediately applicable, but if a second point is found on the line, then the requisite distance is given by $∥\mathbf{A}\times \mathbf{B}∥$ / $∥\mathbf{A}∥$, where $\mathbf{A}\=\mathbf{R}-\mathbf{Q}$ and $\mathbf{B}\=\mathbf{P}-\mathbf{Q}$. Taking $t\=0$ and $t\=1$ on the given line provides the two points Q:$\left(1\,-1\,2\right)$ and R:$\left(8\,-10\,7\right)$, respectively. Hence,

$\mathbf{A}\=\mathbf{R}-\mathbf{Q}\=\left[\begin{array}{c}8\\ -10\\ 7\end{array}\right]-\left[\begin{array}{c}1\\ -1\\ 2\end{array}\right]\=\left[\begin{array}{c}7\\ -9\\ 5\end{array}\right]$ and $\mathbf{B}\=\mathbf{P}-\mathbf{Q}\=\left[\begin{array}{c}2\\ 1\\ -3\end{array}\right]-\left[\begin{array}{c}1\\ -1\\ 2\end{array}\right]\=\left[\begin{array}{c}1\\ 2\\ -5\end{array}\right]$ 

and $\mathbf{A}\times \mathbf{B}\= \|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 7& -9& 5\\ 1& 2& -5\end{array}\|\=\left[\begin{array}{c}35\\ 40\\ 23\end{array}\right]$, so $d\=\frac{∥\mathbf{A}\times \mathbf{B}∥}{\∥\mathbf{A}\∥}$ = $\frac{\sqrt{1225\+1600\+529}}{\sqrt{49\+81\+25}}$ = $\frac{\sqrt{519870}}{155}$

The traditional approach to this calculation is vector-based. The distance is the magnitude of the orthogonal component of a vector U, to point P from an arbitrary point on the line, projected onto V. The arbitrary point on the line is generally taken as A.

The  task template provides the vector ${\mathbf{U}}_{\perp \mathbf{V}}$ , the component of U orthogonal to V.

The distance is then the magnitude of the orthogonal component.

Note that the ProjectionPlot command upon which the task template is based, really does select an arbitrary point on the line. Hence, the figure appearing in the task template differs slightly from Figure 1.6.4(a) where the "arbitrary" point is taken as point A.

A solution with the "Lines & Planes" tools in the Student MultivariateCalculus package:

A traditional vector-based solution: