jmc

Let the line be $$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\mathbf{a}+t\mathbf{d}.$$Then from the given information, $$\begin{array}{r}\left(\begin{array}{c}1\\ 3\\ 8\end{array}\right)=\mathbf{a}-\mathbf{d},\\ \left(\begin{array}{c}0\\ -2\\ -4\end{array}\right)=\mathbf{a}+2\mathbf{d}.\end{array}$$We can treat this system as a linear set of equations in $\mathbf{a}$ and $\mathbf{d}.$ Accordingly, we can solve to get $\mathbf{a}=$ and $\mathbf{d}=$ Hence, $$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}2/3\\ 4/3\\ 4\end{array}\right)+t\left(\begin{array}{c}-1/3\\ -5/3\\ -4\end{array}\right).$$Taking $t=3,$ we get $$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}2/3\\ 4/3\\ 4\end{array}\right)+3\left(\begin{array}{c}-1/3\\ -5/3\\ -4\end{array}\right)=\boxed{\left(\begin{array}{c}-1/3\\ -11/3\\ -8\end{array}\right)}.$$