Three Dimensional Geometry Part 3

L1:‌

x−1

λ

=‌

y−2

1

=‌

z−3

2

‌, ‌
through a point

→

a

1≡(1,2,3)
parallel to

→

b1

≡(λ,1,2)
L2:‌

x+26

−2

=‌

y+18

3

=‌

z+28

λ

through a point

→

a

2=(−26,−18,−28)
parallel to

→

b2

=(−2,3,1)
If lines are coplanar then, (

→

a

2−

→

a

1)⋅

→

b1

×

→

b2

=0
If lines are coplanar then, (

→

a2

−

→

a1

)⋅

→

b1

×

→

b2

=0
⇒|

27	20	31
λ	1	2
−2	3	λ

|=0⇒λ=3
Vector normal to the required plane

→

n

=

→

b1

×

→

b2

⇒

→

n

=|

^ i	^ j	^ k
3	1	2
−2	3	3

|=−3

^

i

−13

^

j

+11

^

k

Equation of plane
≡((x−1),(y−2),(z−3))⋅(−3,−13,11)=0
⇒3x+13y−11z+4=0
Checking the option gives (0,4,5) does not lie on the plane.