Answer
\[
Q(0,4,-2) \text { and } P(4,0,6)
\]
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
\[
\begin{aligned}
&\left\langle x_{0}, y_{0}\right\rangle+t\langle a, b\rangle =\langle x, y\rangle \\
&\left\langle x_{0}, y_{0}, z_{0}\right\rangle+t\langle a, b, c\rangle=\langle x, y, z\rangle
\end{aligned}
\]
It's given the parametric equation of the line is:
\[
2 t=z, \quad -t+3=y, \quad t+1=x
\]
Replace this in our cylinder equation to get:
\[
\begin{aligned}
16=y^{2}+x^{2} & \Rightarrow16=(-t+3)^{2}+(t+1)^{2}\\
& \Rightarrow 16=t^{2}-6 t+9 +t^{2}+2 t+1\\
& \Rightarrow 0=-4 t-6+2 t^{2} \\
& \Rightarrow 0=2\left(-2 t-3+t^{2}\right)\\
& \Rightarrow 0=(1+t)(-3+t) \\
& \Rightarrow -1=t \quad \text { or } \quad 3=t
\end{aligned}
\]
For these two values of the parameter, we have that:
\[
\begin{array}{l}
z_{2}=2(-1)=-2 , y_{2}=3-(-1)=4, \quad x_{2}=1-1=0\\
z_{1}=2(3)=6 , y_{1}=3-3=0, \quad x_{1}=1+3=4
\end{array}
\]
So, the line intersects the cylinder at:
\[
Q(0,4,-2) \text { and } P(4,0,6)
\]
