91Ó°ÊÓ

To find the normal vector to the surface, first, we need to find the gradient of the given function. Let's call the function F(x, y, z): $$F(x, y, z) = \sin x y z - \frac{1}{2} = 0$$. Now, we find the gradient as follows: $$\nabla F(x, y, z) = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right)$$ Start by finding the partial derivative with respect to x: $$\frac{\partial F}{\partial x} = yz\cos x y z$$ Now, find the partial derivative with respect to y: $$\frac{\partial F}{\partial y} = xz\cos x y z$$ Finally, find the partial derivative with respect to z: $$\frac{\partial F}{\partial z} = xy\cos x y z$$ Thus, the gradient vector is: $$\nabla F(x, y, z) = \left(yz\cos x y z, xz\cos x y z, xy\cos x y z \right)$$

Now, we need to evaluate the gradient at the given point: $$\nabla F(\pi, 1, \frac{1}{6}) = \left(\frac{1}{6}\cos{\pi}, \frac{\pi}{6}\cos{\pi},\pi\cos{\pi}\right)$$ The cosine of \(\pi\) is -1, so the gradient is: $$\nabla F(\pi, 1, \frac{1}{6}) = \left(-\frac{1}{6}, -\frac{\pi}{6},-\pi\right)$$ This gradient vector is the normal vector to the tangent plane.

Now, we can use the point-slope form of the equation of a plane with the given point and normal vector: $$A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$$ Using the normal vector components as coefficients and the given point \((\pi, 1, \frac{1}{6})\) as the point on the plane, we get the equation of the tangent plane: $$-\frac{1}{6}(x-\pi) -\frac{\pi}{6}(y-1) -\pi(z-\frac{1}{6}) = 0$$