Find parametric equations for the line tangent to the curve of intersection of the surfaces at the given point.
Surfaces: $\quad x+y^{2}+2 z=7, \quad x=0$
Point: $\quad(0,1,3)$
Find the equations for the tangent line. Let $z=3-2 t$
$x=$
(Type an expression using $\mathrm{t}$ as the variable.)

Step 1 ：The given surfaces are \(x + y^{2} + 2z = 7\) and \(x = 0\). The point of tangency is \((0,1,3)\).

Step 2 ：We can express the curve of intersection of the surfaces as a function of a single variable by substituting \(x = 0\) into the first equation to get \(y^{2} + 2z = 7\).

Step 3 ：Solving this equation for \(y\) in terms of \(z\), we get two solutions: \(y = -\sqrt{7 - 2z}\) and \(y = \sqrt{7 - 2z}\).

Step 4 ：The derivative of \(y\) with respect to \(z\) is \(\frac{1}{\sqrt{7 - 2z}}\).

Step 5 ：The slope of the tangent line is the derivative at the given point, which is 1.

Step 6 ：The equation of the tangent line in point-slope form is \(y = x + 1\).

Step 7 ：We can express this equation in parametric form by letting \(x = t\), which gives \(y = t + 1\).

Step 8 ：We also know that \(z = 3 - 2t\) from the given point.

Step 9 ：Thus, the parametric equations for the tangent line are \(x = t\), \(y = t + 1\), and \(z = 3 - 2t\).

Step 10 ：\(\boxed{x = t, y = t + 1, z = 3 - 2t}\) is the final answer.