$\mathbf{A}\=\mathbf{R}-\mathbf{Q}$ = $\left[\begin{array}{c}5\\ 4\\ 7\end{array}\right]-\left[\begin{array}{c}1\\ 2\\ -3\end{array}\right]$ = $\left[\begin{array}{c}4\\ 2\\ 10\end{array}\right]$ 

$\mathbf{B}\=\mathbf{S}-\mathbf{Q}$ = $\left[\begin{array}{c}6\\ -5\\ -1\end{array}\right]-\left[\begin{array}{c}1\\ 2\\ -3\end{array}\right]$ = $\left[\begin{array}{c}5\\ -7\\ 2\end{array}\right]$ 

$\mathbf{C}\=\mathbf{P}-\mathbf{Q}$ = $\left[\begin{array}{c}2\\ -3\\ 4\end{array}\right]-\left[\begin{array}{c}1\\ 2\\ -3\end{array}\right]$ = $\left[\begin{array}{c}1\\ -5\\ 7\end{array}\right]$

$\left[\mathbf{ABC}\right]$ = $\|\begin{array}{ccc}4& 2& 10\\ 5& -7& 2\\ 1& -5& 7\end{array}\|$ = $-402$

$\mathbf{A}\times \mathbf{B}$ = $\|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 4& 2& 10\\ 5& -7& 2\end{array}\|$ = $\left[\begin{array}{c}74\\ 42\\ -38\end{array}\right]$$$

Initialize

Loading Student:-MultivariateCalculus

Define the position vectors P, Q, R, and S

$⟨2\,-3\,4⟩$$\stackrel{\text{assign to a name}}{\to }$$P$

$⟨1\,2\,-3⟩$$\stackrel{\text{assign to a name}}{\to }$$Q$

$⟨5\,4\,7⟩$$\stackrel{\text{assign to a name}}{\to }$$R$

$⟨6\,-5\,-1⟩$$\stackrel{\text{assign to a name}}{\to }$$S$

By subtraction, obtain the vectors A, B, and C

$\mathbf{A}\=\mathbf{R}-\mathbf{Q}$$\stackrel{\text{assign}}{\to }$

$\mathbf{B}\=\mathbf{S}-\mathbf{Q}$$\stackrel{\text{assign}}{\to }$

$\mathbf{C}\=\mathbf{P}-\mathbf{Q}$$\stackrel{\text{assign}}{\to }$

Apply the appropriate distance formula from Table 1.5.1 

$\|\frac{\mathbf{A}·\left(\mathbf{B}\times \mathbf{C}\right)}{∥\mathbf{A}\times \mathbf{B}∥}\|$ = $\frac{201\sqrt{2171}}{2171}$$\stackrel{\text{at 5 digits}}{\to }$$4.3139$$$

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{MultivariateCalculus}\right)\:$

$\mathbf{P}\,\mathbf{Q}\,\mathbf{R}\,\mathbf{S}≔⟨2\,-3\,4⟩\,⟨1\,2\,-3⟩\,⟨5\,4\,7⟩\,⟨6\,-5\,-1⟩\:$

$\mathbf{A}\,\mathbf{B}\,\mathbf{C}≔\mathbf{R}-\mathbf{Q}\,\mathbf{S}-\mathbf{Q}\,\mathbf{P}-\mathbf{Q}\:$

$\mathrm{abs}\left(\mathrm{BoxProduct}\left(\mathbf{A}\,\mathbf{B}\,\mathbf{C}\right)\right)\/\mathrm{Norm}\left(\mathrm{CrossProduct}\left(\mathbf{A}\,\mathbf{B}\right)\right)$ = $\frac{201\sqrt{2171}}{2171}$$$

Solution from first principles

$d≔\frac{\mathrm{abs}\left(\mathrm{DotProduct}\left(\mathbf{A}\,\mathrm{CrossProduct}\left(\mathbf{B}\,\mathbf{C}\right)\right)\right)}{\mathrm{Norm}\left(\mathrm{CrossProduct}\left(\mathbf{A}\,\mathbf{B}\right)\right)}$ = $\frac{201}{2171}\sqrt{2171}$$$

$\mathrm{evalf}\left(d\right)$ = $4.313860984$$$