Find parametric equations and symmetric equations for the line. The line

Find parametric equations and symmetric equations for the line.
The line through (-6, 2, 3) and parallel to the line $$\displaystyle{\frac{{{1}}}{{{2}{x}}}}={\frac{{{1}}}{{{3}{y}}}}={z}+{1}$$

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Mike Henson

Rationalization. Given a dispacement vector $$\displaystyle{v}={<}{a},{b},{c}{>}$$ which is the difference of the initial position vector $$\displaystyle{r}_{{0}}$$ and final position vector r as well as the value of an initial position vector $$\displaystyle{r}_{{0}}$$, we can acquire the parametric equations of a line:
$\begin{cases}x=x_0+at\\y=y_0+bt\\z=z_0+ct\end{cases}$
From this parametric equation, we can acquire the parametric equations of L as follows:
$$\displaystyle{\frac{{{x}-{x}_{{0}}}}{{{a}}}}={\frac{{{y}-{y}_{{0}}}}{{{b}}}}={\frac{{{z}-{z}_{{0}}}}{{{c}}}}$$
Deriving the parametric equations: finding displacement vector.
$$\displaystyle{\frac{{{1}}}{{{2}}}}{x}+{\frac{{{1}}}{{{3}}}}{y}={z}+{1}$$
This is already a symmetric equation of a line parallel to the given line that passes through (-6,2,3) with position vector $$\displaystyle{r}_{{0}}={<}-{6},{2},{3}{>}$$. We can rewrite this as a parametric equation by equating to t:
$$t=\frac{1}{2}x=\frac{1}{3}u=z+1\Rightarrow\begin{cases}t=\frac{1}{2}x\\t=\frac{1}{3}y\\t=z+1\end{cases}\Rightarrow\begin{cases}x=2t\\y=3t\\z=t-1\end{cases}$$
Deriving the symmetric equations.The symmetric equations are the three-fold equaliy we acquire between x, y and z by eliminating t. It was already mentioned in the chapter that a shorthand formula for this from the parametric equations is:
$$\displaystyle{\frac{{{x}-{x}_{{0}}}}{{{a}}}}={\frac{{{y}-{y}_{{0}}}}{{{b}}}}={\frac{{{z}-{z}_{{0}}}}{{{c}}}}$$
Substituting calues, we get:
$$\displaystyle{\frac{{{x}+{6}}}{{{2}}}}={\frac{{{y}-{2}}}{{{3}}}}={\frac{{{z}-{3}}}{{{1}}}}={\frac{{{x}+{6}}}{{{2}}}}={\frac{{{y}-{2}}}{{{3}}}}={z}-{3}$$