Sow that `x_1 hat i+y_1 hat j+z_1 hat k ,x_2 hat i+y_2 hat j+z_2 hat k ,a n dx_3 hat i+y_3 hat j+z_3 hat k ,` are non-coplanar if `|x_1|>|y_1|+|z_1|,|y_2|>|x_2|+|z_2|a n d|z_3|>|x_3|+|y_3|` .

Explanation

If the given vectors are coplanar, then
`" "|{:(x_1,,y_1,,z_1),(x_2,,y_2,,z_2), (x_3,,y_3,,z_3):}|=0`
or the set of equations
`" "x_1x+y_1y+ z_1z=0`,
and `x_2x+y_2y+z_2z=0`
`" "x_3x+y_3y +z_3z=0` has a non-trivial solution.
Let the given set has a non-trival solution x, y, z without the loss of generality, we can assume that `x ge y ge z`.
For the given equation `x_1x+ y_1y+ z_1z=0`, we have `x_1x= -y_1y- z_1z`. Therefore,
`" "|x_1x|= |y_1y+ z_1z|le |y_1y|+ |z_1z|`
`rArr " "|x_1x| le |y_1x|+ |z_1x|`
`rArr " "|x_1| le |y_1|+ |z_1|`,
which is a contraction to the given inequility, i.e.,
`" "|x_1| gt |y_1|+ |z_1|`,
Similarly, the other inequalities rule out the possibility of a non-trival solution.
Hence, the given equation has only a trival solution.
Hence, the given vectors are non-coplanar.