<s>[INST] Find the point of intersection of the line and surface I have an odd problem with no solution. I am completely lost on how to solve this.
Problem:
Find the coordinates of the point(s) of intersection of the line $x = 1+t$, $y = 2+3t$, $z = 1-t$ and the surface $z = x^2 +2y^2$
Attempt:
$(1) \ x = 1+t$
$(2) \ y = 2+3t$
$(3)\ z = 1-t$
Subbing in $(1), (2), (3)$ into the surface I have
$1-t = (1+t)^2 +2(2+3t)^2$
Solving I get:
$19t^2+27t+8=0$
At this point I need to solve for $t$
It appears to be a quadratic equation.
\begin{array}{*{20}c} {t = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}
\begin{array}{*{20}c} {t = \frac{{ - 27 \pm \sqrt {(27)^2 - 4(19)(8)} }}{{2(19)}}} \end{array}
\begin{array}{*{20}c} {t = \frac{{ - 27 \pm \sqrt {121} }}{{38}}} \end{array}
\begin{array}{*{20}c} \frac{{ - 27 \pm 11 }}{{38}} \end{array}
and 
\begin{array}{*{20}c}\frac{{ - 27 + 11 }}{{38}} and \frac{{ - 27 - 11 }}{{38}} \end{array}
\begin{array}{*{20}c} t= \frac{{ 8 }}{{19}}  \end{array} and 
\begin{array}{*{20}c} t= -1  \end{array}
Then just plug these $t$ values into the parametric equations to get the points?
$(1) \ x = 1+t$ \begin{array}{*{20}c} x= 1 + \frac{{ 8 }}{{19}}  \end{array}
$(2) \ y = 2+3t$ \begin{array}{*{20}c} x= 2 + 3 \frac{{ 8 }}{{19}}  \end{array}
$(3) \ z = 1-t$ \begin{array}{*{20}c} x= 1 - \frac{{ 8 }}{{19}}  \end{array}
Do the same for $t= -1$?
for the points $( 27/19, 26/19, 11/19)$  and $( 0, -1,2)$
Is this correct?
 [/INST] Your method is entirely correct, but as noted in the comment above, you may pick up unwanted solutions. Your point $(0, -1, 2)$ is on the surface, but this is not the case for $(27/19, 26/19, 11/19)$. (This point doesn't satisfy the equation of the surface. So if you double check your computations, and it results in the same coordinates, you can "throw it out," and keep $(0, -1, 2)$.
ADDED The root $t =\dfrac{8}{19}$ should be $t = \left(-\dfrac{8}{19}\right)$. Try computing the coordinates with the negative root $t = -\dfrac{8}{19}$, and see what this brings. Perhaps the point with the modified coordinates will satisfy the equation of the surface!
For $t = \left(-\dfrac{8}{19}\right)$, I get the point $\left(\dfrac{11}{19}, \dfrac{14}{19}, \dfrac{27}{19}\right)$. We need to simply check if it satisfies the equation of the surface. And so it does! (See another confirmation, below, thanks to @RecklessReckoner). 
Therefore, we have two solutions!
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