Answer
\[
t=y, 0=z, t=x
\]
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
\[
b t+y_{0}=y, \quad a t+x_{0}=x
\]
(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
\[
c t+z_{0}=z, \quad b t+y_{0}=y, \quad a t+x_{0}=x
\]
The vector equations of 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 $\langle 1,1,0\rangle=\langle a, b, c\rangle=\vec{v}=$ is parallel to the line
\[
t=x, \quad t-1=y, \quad 2=z
\]
So, this vector is parallel to both lines. Since our line passes through the original $P(a, b, c)=P(0,0,0)$, the parametric equation for the line can be written as follows:
\[
\begin{array}{l}
t+0=y, 0=z, t+0=x \\
t=y, 0=z, t=x
\end{array}
\]
