Answer
\[
2 t-2=x, -t=y, \quad 2 t+5=z
\]
Work Step by Step
(a) The line in 2d space that passes through the point $P_{0}\left(x_{0}, y_{0}\right)$ and that is parallel to the non-zero vector $\mathbf{v}=\langle a, b\rangle=a \mathbf{i}+b \mathbf{j}$ has parametric equations:
\[
a t+x_{0}=x, \quad b t+y_{0}=y
\]
(b) The line in 3d space that passes through the point $P_{0}\left(x_{0}, y_{0}, z_{0}\right)$ and that is parallel to the non-zero vector $\mathbf{v}=\langle a, b, c\rangle=a \mathbf{i}+b \mathbf{j}+c \mathbf{k}$ has parametric equations
\[
a t+x_{0}=x, \quad b t+y_{0}=y, \quad c t+z_{0}=z
\]
The vector equations for these lines can be written as:
\[
\begin{aligned}
&\left\langle x_{0}+a t, y_{0}+b t\right\rangle=\langle x, y\rangle \\
&\left\langle x_{0}+a t, y_{0}+b t, z_{0}+c t\right\rangle=\langle x, y, z\rangle
\end{aligned}
\]
or
\[
\begin{aligned}
&\left\langle x_{0}, y_{0}\right\rangle+t\langle a, b\rangle = \langle x, y\rangle \\
&\left\langle x_{0}, y_{0}, z_{0}\right\rangle+t\langle a, b, c\rangle=\langle x, y, z\rangle
\end{aligned}
\]
Notice that $\langle 2,-1,2\rangle=\vec{v}$ is parallel to the line $2 t+1=x, -t+4=y, z=$ $2 t+6$
So, this vector $\vec{v}=\langle a, b, c\rangle=\langle 2,-1,2\rangle$ is also parallel to the line that passes through $P(-2,0,5)=P\left(x_{0}, y_{0}, z_{0}\right)$ So, the parametric equation of the line can be written as:
\[
\begin{aligned}
x_{0}+a t=x, y_{0}+b t=y, & z_{0}+c t=z \\
\Rightarrow \quad -2+2 t=x, -t=y, & 5+2 t=z
\end{aligned}
\]
