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**Opalg** I'm not sure how to help, because I don't know what method you used to get the tangent plane equation. What I would do would be to start with the point $\displaystyle (x,y,z) = \bigl((2+\cos\phi)\cos\theta, (2+\cos\phi)\sin\theta, \sin\phi\bigr)$ on the torus, noticing that the given point $\displaystyle (1+\frac{\sqrt{3}}{4},\sqrt{3}+\frac{3}{4},\frac{1 }{2})$ corresponds to taking $\displaystyle \phi = \pi/6$ and $\displaystyle \theta = \pi/3$. Then I took the partial $\displaystyle \frac{\partial}{\partial\phi}$ and $\displaystyle \frac{\partial}{\partial\theta}$ partial derivatives of $\displaystyle (x,y,z)$, getting $\displaystyle \bigl(-\sin\phi\cos\theta, -\sin\phi\sin\theta, \cos\phi\bigr)$ and $\displaystyle \bigl(-(2+\cos\phi)\sin\theta, (2+\cos\phi)\cos\theta, 0\bigr)$ as the tangent vectors in the $\displaystyle \phi$ and $\displaystyle \theta$ directions respectively. Next, I formed the cross product of the two tangent vectors to get a normal vector to the tangent plane. After cancelling a factor of $\displaystyle (2+\cos\phi)$, I found that the normal vector was $\displaystyle \bigl(\cos\phi\cos\theta, \cos\phi\sin\theta, \sin\phi\bigr)$. Plugging in the values $\displaystyle \phi = \pi/6$ and $\displaystyle \theta = \pi/3$ gives the normal vector $\displaystyle (\sqrt3,3,2)$. So the equation of the tangent plane at that point is $\displaystyle \sqrt3x + 3y +2z = $ const. Finally, substituting in the values $\displaystyle (x,y,z) = (1+\frac{\sqrt{3}}{4},\sqrt{3}+\frac{3}{4},\frac{1 }{2})$, I got the value $\displaystyle 4+4\sqrt3$ for the constant. So my equation for the tangent plane is $\displaystyle \boxed{\sqrt3x + 3y +2z = 4(1+\sqrt3)}.$ But don't take that on trust, because I could well have made mistakes in the calculations.