How do you determine if the vectors are coplanar: $a=\left[ -2,-1,4 \right],b=\left[ 5,-2,5 \right]$ and $c=\left[ 3,0,-1 \right]$ ?
(a) Cross product them with each other
(b) Finding their scalar triple product
(c) One of them passing through the origin
(d) None of these

Hint: To start with, we are to determine if the vectors given in this problem are coplanar or not. We are to analyze the result by finding their scalar triple product. If the product is valued zero, then they are coplanar and otherwise not coplanar. We will solve the problem by forming a determinant with the components and then finding its value.

Complete step by step answer:
Here in this problem we are provided with three vectors and we are to check if they are coplanar or not.
The three given vectors are, $a=\left[ -2,-1,4 \right],b=\left[ 5,-2,5 \right]$ and $c=\left[ 3,0,-1 \right]$.
Now, to find if they are coplanar or not, we have to find their scalar triple product. If the value of the scalar triple product turns out to be zero, then the vectors are coplanar. On the other hand, if the value is not equal to zero, then we will conclude that the vectors are not coplanar.
An easy way to find the scalar triple product is to evaluate the determinant made by the components one by one and find the value of the determinant. Similarly, if the value of the determinant is said to be zero, then we will get to know that the vectors are coplanar. Again, on the other hand, if the value is said not to be zero, then we will know that the vectors are not coplanar.
Now, to find the determinant,
$\left| \begin{matrix}
   -2 \\
   5 \\
   3 \\
\end{matrix}\begin{matrix}
   -1 \\
   -2 \\
   0 \\
\end{matrix}\begin{matrix}
   4 \\
   5 \\
   -1 \\
\end{matrix} \right|$
Simplifying,
$\Rightarrow -2\left| \begin{matrix}
   -2 & 5 \\
   0 & -1 \\
\end{matrix} \right|-1\left| \begin{matrix}
   5 & 5 \\
   -1 & 3 \\
\end{matrix} \right|+4\left| \begin{matrix}
   5 & -2 \\
   3 & 0 \\
\end{matrix} \right|$
$\Rightarrow -2\left( 2-0 \right)-1\left( 15+5 \right)+4\left( 0-\left( -6 \right) \right)$
$\Rightarrow -2.2-1.20+4.6$
So, we get,
$\Rightarrow -4-20+24=0$
We can see the value of the determinant is zero, then the vectors are said to be coplanar.

Hence the solution is, (b) Finding their scalar triple product .

Note:
The scalar triple product is also obviously very useful if you have a lot of parallelepipeds lying around and want to know their volume. But, if you don't happen to find yourself pining to know the volume of a parallelepiped, you may wonder what's the use of the scalar triple product. Nonetheless, the scalar triple product does have its uses even if you aren't that excited about parallelepipeds. In multivariable calculus, it turns out there are parallelepipeds lurking behind some important formulas and theorems. The reason stems from the definition of the differentiability of functions.