We have,

Let AB be the line segment joining the points

A(1,2,3) and B(–3, 4, –5).

Let XY plane XY – plane divide line AB.at P(x,y,z) in the ratio k : 1.

Co- ordinate of point P(x,y,z) that divide line segment joining point A(x₁,y₁,z₁) and B(x₂,y₂,z₂) in the ratio m : n is

$=(\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}},\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{\mathrm{m}+\mathrm{n}},\frac{{\mathrm{mz}}_2+{\mathrm{nz}}_1}{\mathrm{m}+\mathrm{n}})$

Here,

m : n = k : 1

A(x₁, y₁,z₁) = (1,2,3)

B (x₂,y₂,z₂) = (–3, 4, –5)

Then the co-ordinate of

$\mathrm{P}(\mathrm{x},\mathrm{y},\mathrm{z})=(\frac{\mathrm{k}\times (-3)+1\left(1\right)}{\mathrm{k}+1},\frac{\mathrm{k}\times \left(4\right)+1\times \left(2\right)}{\mathrm{k}+1},\frac{\mathrm{k}\times (-5)+1\times \left(3\right)}{\mathrm{k}+1})$

$=(\frac{-3\mathrm{k}+1}{\mathrm{k}+1},\frac{4\mathrm{k}+2}{\mathrm{k}+1},\frac{-5\mathrm{k}+3}{\mathrm{k}+1})$

Since, point P(x,y,z) lie on the XY –plane

Then, Its Z-coordinate will be zero.

$\mathrm{P}(\mathrm{x},\mathrm{y},0)=(\frac{-3\mathrm{k}+1}{\mathrm{k}+1},\frac{4\mathrm{k}+2}{\mathrm{k}+1},\frac{-5\mathrm{k}+3}{\mathrm{k}+1})$

Then, Comparing and we get,

$\frac{-5\mathrm{k}+3}{\mathrm{k}+1}=0$

– 5k + 3 = 0

– 5k = –3

$\mathrm{k}=\frac{3}{5}$

Then, k : 1 = 3 : 5 

We can substitute the value of k in (1), we get,

$=(\frac{-0.8}{1.6},\frac{4.4}{1.6},0)$

= (–0.5, 2.75,0) 

We get the positive vector of the point of division from (2). Thus, the vector is given by $-0.5\widehat{\mathrm{i}}+2.75\widehat{\mathrm{j}}$. Here, $\widehat{\mathrm{i}}$ denotes unit vector to X-axis and $\widehat{\mathrm{j}}$ denotes unit vector to Y-axis.

Hence, this is the answer.