The wire connects points \( A(4, 1, -2) \) and \( B(6, 2, -3) \). Lanterns are positioned at points that divide the wire into three equal segments. These points, \( P_1 \) and \( P_2 \), trisect the line segment AB. \( P_1 \) is located closer to A, dividing the segment in a \( 1:2 \) ratio, while \( P_2 \) is closer to B, dividing the segment in a \( 2:1 \) ratio. The section formula is used to find these coordinates: \( x = \frac{m x_2 + n x_1}{m + n}, \quad y = \frac{m y_2 + n y_1}{m + n}, \quad z = \frac{m z_2 + n z_1}{m + n} \).
1. For \( P_1 \) (ratio \( 1:2 \), \( m=1, n=2 \)):
\( x_1 = \frac{1 \cdot 6 + 2 \cdot 4}{1 + 2} = \frac{14}{3} \)
\( y_1 = \frac{1 \cdot 2 + 2 \cdot 1}{1 + 2} = \frac{4}{3} \)
\( z_1 = \frac{1 \cdot (-3) + 2 \cdot (-2)}{1 + 2} = \frac{-7}{3} \)
Thus, \( P_1 \left( \frac{14}{3}, \frac{4}{3}, \frac{-7}{3} \right) \).
2. For \( P_2 \) (ratio \( 2:1 \), \( m=2, n=1 \)):
\( x_2 = \frac{2 \cdot 6 + 1 \cdot 4}{2 + 1} = \frac{16}{3} \)
\( y_2 = \frac{2 \cdot 2 + 1 \cdot 1}{2 + 1} = \frac{5}{3} \)
\( z_2 = \frac{2 \cdot (-3) + 1 \cdot (-2)}{2 + 1} = \frac{-8}{3} \)
Thus, \( P_2 \left( \frac{16}{3}, \frac{5}{3}, \frac{-8}{3} \right) \).
The lantern positions are \( P_1 \left( \frac{14}{3}, \frac{4}{3}, \frac{-7}{3} \right) \) and \( P_2 \left( \frac{16}{3}, \frac{5}{3}, \frac{-8}{3} \right) \).