Answer
\[
P(1,-1,2)
\]
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
\[
\begin{aligned}
&\left\langle x_{0}, y_{0}, z_{0}\right\rangle+t\langle a, b, c\rangle= \langle x, y, z\rangle \\
&\left\langle x_{0}, y_{0}\right\rangle+t\langle a, b\rangle =\langle x, y\rangle
\end{aligned}
\]
There are parametric equations for two lines in 3d space:
\[
\begin{array}{ll}
2+t=x, & 2+3 t=y, \quad 3+t=z \\
2+t=x, & 3+4 t=y \quad, \quad 4+2 t=z
\end{array}
\]
These lines are intercepted if both pass through the point $P\left(x_{0}, y_{0}, z_{0}\right)$. If we equate our y-coordinates, we have that $2+3 t=3+4 t \Rightarrow t=2-3=-1$ For this value of the parameter ,we have that for $L_{1}:$
\[
P=P(2-1,2+3(-1), 3-1)=(1,-1,2)P
\]
and for $L_{2}$
\[
P=P(2-1,3+4(-1), 4+2(-1))=(1,-1,2)P
\]
So, both lines are intercepted at $(1,-1,2)P$
