Answer
The line intersects the plane at $P(5 / 4,9 / 4,1 / 2)$
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
\[
c t+z_{0}=z, \quad b t+y_{0}=y, \quad at+x_{0}=x
\]
The vector equations of these lines can be written as:
\[
\begin{aligned}
&\left\langle x_{0}+a t, y_{0}+b t, z_{0}+c t\right\rangle =\langle x, y, z\rangle \\
&\left\langle x_{0}+a t, y_{0}+b t\right\rangle =\langle x, y\rangle
\end{aligned}
\]
or
\[
\begin{aligned}
&\left\langle x_{0}, y_{0}, z_{0}\right\rangle+t\langle a, b, c\rangle=\langle x, y, z\rangle \\
&\left\langle x_{0}, y_{0}\right\rangle+t\langle a, b\rangle =\langle x, y\rangle
\end{aligned}
\]
It's given the parametric equation of the line is:
\[
2 t-1=z, \quad 3 t=y \quad, \quad 2-t+2=x
\]
Replace this in the equation of the plane to get:
\[
\begin{aligned}
6=3 z+2 y & \Rightarrow 6=(2 t-1)3+(3 t)2 \\
& \Rightarrow 6= 6 t+6 t-3\\
& \Rightarrow 9=12 t \\
& \Rightarrow 3 / 4=9 / 12=t
\end{aligned}
\]
For this value of $f$, we get:
\[
\begin{array}{l}
z=-1+2 t=-1+\frac{3}{2}=\frac{1}{2} \\
y=3 t=3(3 / 4)=9 / 4 \\
x=2-t=2-3 / 4=5 / 4
\end{array}
\]
This means that the line intersects the plane at $P(5 / 4,9 / 4,1 / 2)$
