(a)
The parametric equations for a line passing through \((x_{0},\ y_{0},\ z_{0})\) and parallel to the vector
\(v = ai\ +\ bj\ +\ ck\) are
\(x = x_{0}\ +\ at,\ y = y_{0}\ +\ bt,\ z = z_{0}\ +\ ct\)
The required parametric equations are
\(x =\ −1\ +\ 3t,\ y = 4t,\ z = 8\ −\ 8t\)
(b)
The parametric equations are
\(x =\ −1\ +\ 3t,\ y = 4t,\ z = 8\ −\ 8t\)
Solving for t we have,
\(\Rightarrow\ t=\ \frac{x\ +\ 1}{3},\ t=\ \frac{y}{4},\ t=\ \frac{z\ -\ 8}{-8}\)

\(\Rightarrow\ \frac{x\ +\ 1}{3}=\ \frac{y}{4}=\ \frac{z\ -\ 8}{-8}\)

\(\Rightarrow\ \frac{x\ +\ 1}{3}=\ \frac{y}{4}=\ \frac{8\ -\ z}{8}\) Symmetric equations \(\frac{x\ +\ 1}{3}=\frac{y}{4}=\frac{8\ -\ z}{8}\)

\(\Rightarrow\ \frac{x\ +\ 1}{3}=\ \frac{y}{4}=\ \frac{z\ -\ 8}{-8}\)

\(\Rightarrow\ \frac{x\ +\ 1}{3}=\ \frac{y}{4}=\ \frac{8\ -\ z}{8}\) Symmetric equations \(\frac{x\ +\ 1}{3}=\frac{y}{4}=\frac{8\ -\ z}{8}\)