xy plane divides the line joining the points \[\left( {2,4,5} \right)\] and $\left( { - 4,3, - 2} \right)$ in the ratio
A. $3:5$
B. $5:2$
C. $1:3$
D. $3:4$

Hint: In order to solve this type of question, first assume the ratio as k:1. Then, using the section formula i.e., $\left( {x,y,z} \right) = \left( {\dfrac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}},\dfrac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}},\dfrac{{{m_1}{z_2} + {m_2}{z_1}}}{{{m_1} + {m_2}}}} \right)$ we will find the required ratio. Here, $\left( {{x_1},{y_1},{z_1}} \right) = \left( {2,4,5} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right) = \left( { - 4,3, - 2} \right)$. Also, ${m_1} = k$ and ${m_2} = 1.$

Formula used:
$\left( {x,y,z} \right) = \left( {\dfrac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}},\dfrac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}},\dfrac{{{m_1}{z_2} + {m_2}{z_1}}}{{{m_1} + {m_2}}}} \right)$

Complete step by step solution:
We are given that,
$A\left( {2,4,5} \right)$ and $B\left( { - 4,3, - 2} \right)$
Let xy plane divide the line joining the points in the ratio of $k:1.$
In the xy plane, z-coordinate must be $0.$
Solving for z-coordinate,
Using section formula, compare the z-coordinate to get the required ratio,
$ \Rightarrow \dfrac{{k\left( { - 2} \right) + 1\left( 5 \right)}}{{k + 1}} = 0$
$ - 2k + 5 = 0$
On solving,
$k = \dfrac{5}{2}$
$\therefore $The correct option is B.

Note: The line segment divides the xy plane, this means that the line joining the given points is parallel to z-axis. So, the z-coordinate is 0. Also, make sure that the coordinates of the point that divides the plane is $\left( {x,y,0} \right)$ and not $\left( {0,0,z} \right)$, otherwise it may lead to incorrect answer.