Answer 1

We use the results: `bar"b" xx bar"b" = bar"0"` and if in a scalar triple product, two vectors are equal, then the scalar triple product is zero.

`[bar"a" bar"b" bar"c"] = bar"a".(bar"b" xx bar"c")`

`= bar"a".[bar"b" xx (3bar"a" - 2bar"b")]`

`= bar"a".(3bar"b" xx bar"a" - 2bar"b" xx bar"b")`

`= bar"a". (3bar"b" xx bar"a" - bar"0")`

`= 3bar"a".(bar"b" xx bar"a") = 3 xx 0 = 0`

Alternative Method:

`bar"c" = 3bar"a" - 2bar"b"`

∴ `bar"c"` is a linear combination of `bar"a"  "and"  bar"b"`

∴ `bar"a" , bar"b" , bar"c"` are coplanar

∴ `[bar"a" , bar"b" , bar"c"] = 0.`