Finding the Equation of a Tangent Line to a implicit curve

Question

Let be $C=\{(x,y,z): y=2+e^x\sin(2\pi z), \quad z=y^2-\ln(x+1)-3\}$ at $P=(0,2,1)$. I tried to find it and I get the equation is

$$ z=t,\qquad y=2\pi t-2\pi +2\qquad \text {and}\qquad x=(8\pi-1)t-8\pi+1. $$

I am not sure if I am right. I appreaciate any help and I also want to get the graph in some programm online.

Thank you!

Answer 1

Let $f=y-2-e^x\sin(2\pi z)$ and $g=z-y^2+\ln(x+1)+3$ at $p=(0,2,1)$ we have $$\vec{v}=\nabla f(p)\times\nabla g(p)=(8\pi-1)\vec{i}+2\pi\vec{j}-\vec{k}$$ then $\dfrac{x}{8\pi-1}=\dfrac{y-2}{2\pi}=\dfrac{z-1}{-1}$.