Prove that the four points `6 hat i-7 hat j ,16 hat i-19 hat j-4 hat k ,3 hat j-6 hat ka n d2 hat i+5 hat j+10 hat5` form a tetrahedron in space.

Explanation

Given points are `A(6hati - 7hatj), B(16hati - 19 hatj - 4hatk), C(3hatj - 6hatk), D(2hati + 5hatj + 10 hatk)`
Hence, vectors `vec(AB) = 10 hati - 12hatj -4 hatk, vec(AC) = -6hati + 10 hatj - 6 hatk and vec(AD) = -4hati + 12hatj + 10 hatk`
Now determinant of coefficients of `vec(AB), vec(AC), vec(AD)` is
`" "|{:(10,,-12,,-4),(-6,,10,,-6),(-4,,12,,10):}|= 10 (100+ 72) + 12(- 60 - 24) - 4(-72+ 40) ne 0`
Hence, the given points are non-coplanar and therefore form a tetrahedron in space.