Find parametric equations for the line tangent to the curve of intersection of the surfaces at the given point.
Surfaces: $\quad x+y^{2}+2 z=5, \quad x=0$
Point: $\quad(0,1,2)$
Find the equations for the tangent line. Let $z=2-2 t$.
$x=0$ (Type an expression using $t$ as the variable.)
$y=\square$ (Type an expression using $t$ as the variable.)

Step 1 ：The given surfaces are $x+y^{2}+2 z=5$ and $x=0$. The point of tangency is $(0,1,2)$.

Step 2 ：The curve of intersection of the surfaces is given by $y^{2}+2 z=5$ (since $x=0$).

Step 3 ：To find the tangent line, we need to find the derivative of the curve. Let's write the equation in terms of $y$ and $z$ as $y^{2}=5-2z$.

Step 4 ：Differentiating both sides with respect to $z$, we get $2y \frac{dy}{dz}=-2$.

Step 5 ：Solving for $\frac{dy}{dz}$, we get $\frac{dy}{dz}=-\frac{1}{y}$.

Step 6 ：Substituting the point $(0,1,2)$ into the derivative, we get $\frac{dy}{dz}=-1$ at the point of tangency.

Step 7 ：The slope of the tangent line is $-1$. So, the equation of the tangent line can be written in parametric form as $y=1-t$ (since the slope is $-1$).

Step 8 ：Given that $z=2-2t$, the parametric equations for the tangent line are $x=0$, $y=1-t$, and $z=2-2t$.