Vector perpendicular to the plane?

Question

The 2 vectors 3i-7j (no k) and 2i+3j-2k define a plane (its the plane of the triangle with both tails at one vertex and each head at one of the other vertices).
Which of the following vectors is perpendicular to the plane?
a 4i+6j+13k
b -4i+6j+13k
c 4i-6j+13k
d 4i+6j-13k
e 4i+6j

I did the cross product and got
=i(AyBk-AkBy)+j(AkBx-AxBk)+k(AxBy-AyBx)
=14i+6j+23k

How do I go from here?

Answer 1

the cross product is:
( 14,6 ,23 )
and any vector perpendicular to the plane has to be a multiple
of this vector, so my guess is that,
none of the above

Answer 2

For a vector to be perpendicular to the plane, it would have to be parallel to the cross product, and parallel vectors are all scalar multiples of each other. Thus, you see whether any of the listed vectors is actually a scalar multiple of the cross product. In this case, none of them are, so none of the listed vectors are perpendicular to the plane (indeed, none of them are perpendicular to 3i-7j).