Three Dimensional Geometry Part 2

P⇒lx+my+nz=0
P contains L1
L1⇒‌

x−1

−1

=‌

y+4

2

=‌

z+2

3

So, (1,−4,−2) lies on plane.
1−4m−2n=0...(i)
And (−1,2,3) will be perpendicular to (I,m,n).
−1+2m+3n=0....(ii)
Adding Eqs. (i) and (ii),
−2m+n=0
n=2m
l−4m−4m=0
l=8m
So, l=8m and n=2m
Plane ⇒8x+y+2z=0
Now, A(−3,−6,1) and B(2,4,−3)
Plane P divides AB in the ratio of k:1.
Let plane P intersect the line AB at point O.
So, O=(‌

2k−3

k+1

,‌

4k−6

k+1

,‌

−3k+1

k+1

)
And O lies on plane P,
So, 8(2k−3)+(4k−6)+2(−3k+1)=0
⇒‌14k−28‌‌=0
∴‌k‌‌=2