\( 1 \leftarrow \) Find an equation of the line passing through \( P_{0} \) and normal to the plane \( P \). \( P_{0}(1,2,3) ; P: 2 x-3 y+z=10 \) Choose the correct answer below. A. An equation of the line is \( \langle 1-2 t, 2+3 t, 3-t\rangle \). B. An equation of the line is \( \langle 2-t,-3-2 t, 1-3 t\rangle \). An equation of the line is \( \langle 2-t,-3-2 t, 1-3 t\rangle \). D. An equation of the line is \( \langle 1+2 t, 2-3 t, 3+t\rangle \).

To find the equation of the line passing through the point \( P_0(1, 2, 3) \) and normal to the plane defined by \( P: 2x - 3y + z = 10 \), we first identify the normal vector to the plane. The coefficients of \( x, y, z \) in the plane's equation give us the normal vector \( \mathbf{N} = \langle 2, -3, 1 \rangle \). Using the point-normal form of a line, the equation can be expressed as: \[ \langle x, y, z \rangle = \langle 1, 2, 3 \rangle + t \langle 2, -3, 1 \rangle, \] where \( t \) is a parameter. Thus, we find: - \( x = 1 + 2t \) - \( y = 2 - 3t \) - \( z = 3 + t \) This correlates with option D. So, the correct answer is: D. An equation of the line is \( \langle 1+2 t, 2-3 t, 3+t\rangle \).