Find the equation of the plane that is parallel to the vectors \( \langle 2,0,3\rangle \) and \( \langle 0,3,1\rangle \), passing through the point \( (2,0,-3) \). The equation of the plane is (Type an equation using \( x, y \), and \( z \) as the variables.)

To find the equation of the plane, we first need to determine the normal vector to the plane. This can be obtained by taking the cross product of the two given vectors \( \langle 2,0,3\rangle \) and \( \langle 0,3,1\rangle \). Calculating the cross product: \[ \mathbf{n} = \langle 2,0,3\rangle \times \langle 0,3,1\rangle = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 0 & 3 \\ 0 & 3 & 1 \end{vmatrix} \] The determinant expands to: \[ \mathbf{n} = \mathbf{i}(0 \cdot 1 - 3 \cdot 3) - \mathbf{j}(2 \cdot 1 - 0 \cdot 3) + \mathbf{k}(2 \cdot 3 - 0 \cdot 0) \] \[ = \mathbf{i}(0 - 9) - \mathbf{j}(2 - 0) + \mathbf{k}(6 - 0) \] \[ = -9\mathbf{i} - 2\mathbf{j} + 6\mathbf{k} \] Thus, the normal vector is \( \langle -9, -2, 6 \rangle \). Using the point-normal form of the plane's equation, \( n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \), where \( n_1, n_2, n_3 \) are the components of the normal vector and \( (x_0, y_0, z_0) \) is the given point \( (2,0,-3) \): Substituting in the values, we have: \(-9(x - 2) - 2(y - 0) + 6(z + 3) = 0\) Distributing terms gives: \(-9x + 18 - 2y + 6z + 18 = 0\) Combining like terms results in: \(-9x - 2y + 6z + 36 = 0\) Rearranging this leads us to the final equation of the plane: \[ 9x + 2y - 6z = 36 \]