Question

The value of $\left[(a-b)(b-c)(c-a)\right]$ is

• A. $0$
• B. $1$
• C. $2\left[abc\right]$
• D. $2$

Answer 1

The correct option is A

$0$

Explanation for correct option

According to the condition for co-planarity, the sum of given vector $a-b,b-c$ and $c-a$ should be zero.

Given, $\left[(a-b)(b-c)(c-a)\right]$ can be expressed as $\left[xyz\right]$

The sum of this should be $x+y+z=0$

So, $a-b+b-c+c-a$ will also be $0$

Thus, the points are coplanar and the value of $\left[(a-b)(b-c)(c-a)\right]$ is $0$.

Hence, the correct answer is option A.