Make the 3 dimensional determinant with 5 columns so that it is easier to visualize what is being computed:

#<3,0,4>xx<-1,5,2> =|
(hati,hatj,hatk,hati,hatj),
(3,0,4,3,0),
(-1,5,2,-1,5)
|#

Please observe that columns 4 and 5 are copies of columns 1 and 2 respectively.

I will make the elements of the determinant that are used in each computation turn #color(red)(red)#

Compute the major diagonal for the #hati# component:

#|
(color(red)hati,hatj,hatk,hati,hatj),
(3,color(red)0,4,3,0),
(-1,5,color(red)2,-1,5)
| = color(red)((0)(2)hati)#

Subtract the minor diagonal for the #hati# component:

#|
(hati,hatj,hatk,color(red)hati,hatj),
(3,0,color(red)4,3,0),
(-1,color(red)5,2,-1,5)
| = {(0)(2) color(red)(- (4)(5))}color(red)hati#

Add the major diagonal for the #hatj# component:

#|
(hati,color(red)hatj,hatk,hati,hatj),
(3,0,color(red)4,3,0),
(-1,5,2,color(red)(-1),5)
| = {(0)(2) - (4)(5)}hati + color(red)((4)(-1)hatj)#

Subtract the minor diagonal for the #hatj# component:

#|
(hati,hatj,hatk,hati,color(red)hatj),
(3,0,4,color(red)3,0),
(-1,5,color(red)2,-1,5)
| = {(0)(2) - (4)(5)}hati + {(4)(-1)-color(red)((3)(2))}color(red)hatj#

Add the major diagonal for the #hatk# component:

#|
(hati,hatj,color(red)hatk,hati,hatj),
(3,0,4,color(red)3,0),
(-1,5,2,-1,color(red)5)
| = {(0)(2) - (4)(5)}hati + {(4)(-1)-(3)(2)}hatj + color(red)((3)(5)hatk)#

Subtract the minor diagonal for the #hatk# component:

#|
(hati,hatj,color(red)hatk,hati,hatj),
(3,color(red)0,4,3,0),
(color(red)(-1),5,2,-1,5)
| = {(0)(2) - (4)(5)}hati + {(4)(-1)-(3)(2)}hatj + {(3)(5) color(red)(- (0)(-1))}color(red)hatk#

Simplify:

#|
(hati,hatj,hatk,hati,hatj),
(3,0,4,3,0),
(-1,5,2,-1,5)
| = -20hati -10hatj + 15hatk#

You may prefer the form #< -20, -10, 15>#

This vector will be perpendicular to both given vectors but let's compute the dot-product and verify that both dot products are zero:

#< -20, -10, 15>*<3, 0, 4> = (-20)(3)+(-10)(0) + (15)(4) = 0#

#< -20, -10, 15>*<-1, 5, 2> = (-20)(-1)+(-10)(5) + (15)(2) = 0#