Answer
\[
-t+2=x, \quad 2 t-1=y, \quad 7 t+5=z
\]
Work Step by Step
(a) The line in 2d space that passes through the point $P_{0}\left(x_{0}, y_{0}\right)$ and 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 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
\[
a t+x_{0}=x, \quad z_{0}+c t=z, \quad b t+y_{0}=y
\]
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
\[
\begin{aligned}
&\left\langle x_{0}, y_{0}\right\rangle+t\langle a, b\rangle=\langle x, y\rangle \\
&\left\langle x_{0}, y_{0}, z_{0}\right\rangle+t\langle a, b, c\rangle=\langle x, y, z\rangle
\end{aligned}
\]
From this theory, we have the parametric equation for the line that passes through
$P(2,-1,5)$ and that is parallel to $\vec{v}=\langle-1,2,7\rangle$ can be written as:
\[
\begin{aligned}
c t+z_{0}=z , & b t+y_{0}=y, \quad a t+x_{0}=x \\
\Rightarrow \quad -t+2=x, \quad 2 t-1=y, & 7 t+5=z
\end{aligned}
\]
