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I am struggling with finding the parametric equation for this problem. I just started this chapter so I do apologize if this seems like a dumb question.

"Find equations for the tangent plane and parametric equations of the normal line to the given surface at the specified point (Enter your answer as a comma-separated list of equations. Let x, y, and z be in terms of t.) of the following."

Given: $\frac{5}{2}(x − z) = 10arctan(yz)$ , $(1 + 𝜋, 1, 1)$

I have found the tangent plane to be: $\frac{1}{4}\left(x-1-\pi \right)-\frac{1}{2}\left(y-1\right)-\frac{3}{4}\left(z-1\right)=0$

However, for the parametric equation, my answer was $\left(\frac{t}{4}+1+\pi ,-\frac{t}{2}+1,-\frac{3t}{4}+1\right)$, but I keep getting it incorrect.

Any help is appreciated!

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$\nabla (\frac 52(x-z)-10\arctan yz) = (\frac 52,-\frac {10z}{1+y^2z^2},-\frac 52 - \frac {10y}{1+y^2z^2})$
evaluated at $(1+\pi,1,1)$ is $(\frac 52,-5,-\frac {15}{2})$

The normal line: $(x,y,z) = (1+\pi,1,1)+(\frac 52,-5,-\frac {15}{2})t.$ Or,
$x = \frac 52 t + 1+\pi\\ y = -5t+1\\ z = -\frac {15}{2} t + 1$

IF you wanted to scale this up or down, that would be okay.

$x = t + 1+\pi\\ y = -2t+1\\ z = -3t + 1$

And the plane is: $(x-1-\pi) - 2(y-1) - 3(z-1) = 0$

This looks to be equivalent to what you have in the original post.

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