Three Dimensional Geometry Part 1

Vector along the normal to the plane containing thelines

x

3

=

y

4

=

z

2

and

x

4

=

y

2

=

z

3

is (8

^

i

−

^

j

−10

^

k

)
vector perpendicular to the vectors 2

^

i

+3

^

j

+4

^

k

and 8

^

i

−

^

j

−10

^

k

is 26

^

i

−52

^

j

+26

^

k

so, required plane is
26x – 52y + 26z = 0
x – 2y + z = 0