Point A is given as (2,4,5) and Point B is given as (3,5,-4).

Also it is given that YZ Palne divides the line AB. Let the YZ plane divide AB at some point M denoted by (x,y,z) in the ratio k:1.

Let A = (2,4,5) = (x1 , y1 , z1) and B = (3,5,-4) = (x2 , y2 , z2)

Therefore :- x1 = 2 , y1 = 4 , z1 = 5 and x2 = 3 , y2= 5 , z2= -4.

Now , co-ordinate of M which divides the line AB in ratio suppose m:n is given by :-

[ { (mx2 + nx1)/(m+n) }, { (my2 + ny1)/(m+n) } , { (mz2 + nz1)/(m+n) } ]

Since we are considering the ratio as k:1 we subsitute m = k and n = 1 in above equation.

Therefore,

(x, y, z) = [ { (3k + 2)/(k+1) } , { (5k + 4)/(k +1 ) } , { (-4k + 5)/(k+1) } ]

Now since the plane lies in YZ Plane it's x co-ordinate will be zero.

Therefore by taking x = 0 in above equation we get ,

(0, y, z) = [ { (3k + 2)/(k+1) } , { (5k + 4)/(k +1 ) } , { (-4k + 5)/(k+1) } ]

Now by Comparing x co-ordinate on both the sides we get :-

0 = (3k+2)/(k+1)

Therefore k = -2/3.

Since we got k as negative , it indicates that the point divides the line Externally in the ratio of 2 : 3.

If you still have any queries feel free to ask in the comment section down below.

Hii

Given -- A (x,y,z)= (2,4,5)  and B(x,y,z)=(3,5,-4)

YZ plane divide the line giving A (x,y,z) and B(x,y,z)in the ratio  ( -x1: x2)
= -2 : 3
= (l : m)

let point of intersection  =  {(lx2 + m x1) / (l + m)},  {(ly2 + my1)/ (l + m)} ,  {(lz2 + lz1) / (l+m)}
= {-2(3) + 3(2)} / (-1+3) ,  { -2(5)+3(4)} / (-2+3) , { -2(-4) +3(5)}  / (-2+3)
= (-6+6) / 1 , (-10+12) / 1 , (8+15) / 1
= ( 0, 2, 23) ANS

I hope this help. Please put it on paper and then try to understand what is done.

Thanks!!