The tangent to [tex]C[/tex] through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces [tex]S[/tex] and [tex]T[/tex] at that point.
Let [tex]f(x,y,z)=x^2+2y^3+3z^4[/tex]. Then [tex]S[/tex] is the level curve [tex]f(x,y,z)=6[/tex]. Recall that the gradient vector is perpendicular to level curves; we have
[tex]\nabla f(x,y,z)=(2x,6y,12z^2)[/tex]
so that the gradient of [tex]f[/tex] at (1, 1, 1) is
[tex]\nabla f(1,1,1)=(2,6,12)[/tex]
For the surface [tex]T[/tex], we have
[tex]\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0[/tex]
so that [tex]\vec r(1,0)=(1,1,1)[/tex]. We can obtain a vector normal to [tex]T[/tex] by taking the cross product of the partial derivatives of [tex]\vec r(u,v)[/tex], and evaluating that product for [tex]u=1,v=0[/tex]:
[tex]\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left(u-2ve^v,-1,\dfrac{2e^v}u\right)[/tex]
[tex]\left(\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right)(1,0)=(1,-1,2)[/tex]
Now take the cross product of the two normal vectors to [tex]S[/tex] and [tex]T[/tex]:
[tex](2,6,12)\times(1,-1,2)=(24,8,-8)[/tex]
The direction of vector (24, 8, -8) is the direction of the tangent line to [tex]C[/tex] at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by [tex]t\in\Bbb R[/tex]. Then adding (1, 1, 1) shifts this line to the point of tangency on [tex]C[/tex]. So the tangent line has equation
[tex]\vec\ell(t)=(1,1,1)+t(24,8,-8)=(1+24t,1+8t,1-8t)[/tex]
