Calculate the value of the forms \(\omega_{1}=d x_{2} \wedge d x_{3}, \omega_{2}=x_{1} d x_{3} \wedge d x_{2}\), and \(\omega_{3}=d x_{3} \wedge d r^{2}\left(r^{2}=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)\), on the pair of vectors \(\xi=(1,1,1), \eta=(1,2,3)\) at the point \(\mathbf{x}=(2,0,0)\).

We have three given forms, \(\omega_1\), \(\omega_2\), and \(\omega_3\). First, we need to calculate the wedge product for each form. For \(\omega_1\), we have \(d x_{2} \wedge d x_{3}\). The wedge product is linear, and anti-commutative, so we have: \(\omega_{1}=d x_{2} \wedge d x_{3} = -d x_{3} \wedge d x_{2}\) For \(\omega_2\), we have \(x_{1} d x_{3} \wedge d x_{2}\). The wedge product is linear and anti-commutative, so once again we have: \(\omega_{2} = x_{1} d x_{3} \wedge d x_{2} = x_{1} (- d x_{2} \wedge d x_{3}) = - x_{1} d x_{2} \wedge d x_{3}\) Lastly, for \(\omega_3\), we have \(d x_{3} \wedge d r^{2}\). First, we must compute the differential \(d r^{2}\). Given that \(r^{2} = x_{1}^{2} + x_{2}^{2} + x_{3}^{2}\), we have: \(dr^{2} = 2 x_{1} dx_{1} + 2 x_{2} dx_{2} + 2 x_{3} dx_{3}\) Now, we can calculate the wedge product: \(\omega_{3}=d x_{3} \wedge d r^{2} = d x_{3} \wedge (2 x_{1} dx_{1} + 2 x_{2} dx_{2} + 2 x_{3} dx_{3})\)

Now that we have the computed wedge products for each form, we need to plug in the given vectors and point \(\mathbf{x}\) to each form. For \(\omega_{1}\) we have: \(\omega_{1}(\xi, \eta) = -d x_{3}(\xi) \wedge d x_{2}(\eta)\) For \(\omega_{2}\) we have: \(\omega_{2}(\xi, \eta) = - x_{1} d x_{2}(\xi) \wedge d x_{3}(\eta)\) For \(\omega_{3}\) we have: \(\omega_{3}(\xi, \eta) = d x_{3}(\xi) \wedge (2 x_{1} dx_{1}(\eta) + 2 x_{2} dx_{2}(\eta) + 2 x_{3} dx_{3}(\eta))\)

Lastly, we need to evaluate the expressions at the point \(\mathbf{x} = (2, 0, 0)\). For \(\omega_{1}\): \(\omega_{1} = -d x_{3}(\xi) \wedge d x_{2}(\eta) = -1 \wedge 2 = 2\) For \(\omega_{2}\): \(\omega_{2} = - x_{1} d x_{2}(\xi) \wedge d x_{3}(\eta) = -2 (1 \wedge 3) = -6\) For \(\omega_{3}\): \(\omega_{3} = d x_{3}(\xi) \wedge (2 x_{1} dx_{1}(\eta) + 2 x_{2} dx_{2}(\eta) + 2 x_{3} dx_{3}(\eta)) = 1 \wedge (4 + 0 + 0) = 4\) So we have the values of the forms: \(\omega_{1}(\xi, \eta) = 2\) \(\omega_{2}(\xi, \eta) = -6\) \(\omega_{3}(\xi, \eta) = 4\)