The second point should be 2i+3j-4k?
By taking pairs of points we can work out three line vectors. By taking two such lines, we can use the vector cross-product to find the normal orthogonal vector. The third vector line will be orthogonal to the normal and therefore its dot product with the normal vector will be zero.
Call the points P1(3,-2,-1), P2(2,3,-4), P3(-1,1,2), P4(4,5,n) and vectors V1=P2-P1, V2=P3-P1, V3=P4-P1. Note that P1 is taken to be the reference point from which the three vector lines radiate.
V1=-i+5j-3k, V2=-4i+3j+3k, V3=i+7j+(n+1)k
| X-product of unit vectors |
i |
j |
k |
| i |
0 |
k |
-j |
| j |
-k |
0 |
i |
| k |
j |
-i |
0 |
V4=V2 X V3=-28k+4(n+1)j-3k+3(n+1)i+3j-21i=3(n-6)i+(4n+7)j-31k.
V1 . V4=0=3(6-n)+5(4n+7)+93
18-3n+20n+35+93=0
17n+146=0, n=-146/17