 Are the four points A(1,−1,1), B(−1,1,1), C(1,1,1) and D(2,−3,4) coplanar? Justify your answer. Verified
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Hint: Here first we have to find the scalar triple product of three vectors formed by the given four points and then check that value using the condition of coplanar.
We know that condition of coplanar is that “3 vectors are coplanar, if their scalar triple product is zero”

Here, we have the given points as
$A\left( {1, - 1,1} \right)$, $B\left( { - 1,1,1} \right)$, $C\left( {1,1,1} \right)$ and $D\left( {2, - 3,4} \right)$
Now, we consider in position vector of points
The position vectors of the point $A,B,C,D$ are
$\overline a = \widehat i - \widehat j + \widehat k$
$\overline b = - \widehat i + \widehat j + \widehat k$
$\overline c = \widehat i + \widehat j + \widehat k$
$\overline d = 2\widehat i - 3\widehat j + 4\widehat k$
Now we are going to find $\overline {AB}$, $\overline {AC}$ and $\overline {AD}$ using the given four points,
We know that $\overline {AB}$ is found by subtracting $\overline a$ from $\overline b$, that is,
$\overline {AB} = \overline b - \overline a$
$\overline {AB} = \left( { - \widehat i + \widehat j + \widehat k} \right) - \left( {\widehat i - \widehat j + \widehat k} \right) = - 2\widehat i + 2\widehat j$
Using the similar conditions we will find the values of $\overline {AD}$ $and$ $\overline {AC}$
$\overline {AC} = \overline c - \overline a$
$\overline {AC} = \left( {\widehat i + \widehat j + \widehat k} \right) - \left( {\widehat i - \widehat j + \widehat k} \right) = 2\widehat j$
$\overline {AD} = \overline d - \overline a$
$\overline {AD} = \left( {2\widehat i - 3\widehat j + 4\widehat k} \right) - \left( {\widehat i - \widehat j + \widehat k} \right) = \widehat i - 2\widehat j + 3\widehat k$​
Now we have found three vectors using the given four points, our next process is to find the scalar triple product of the three vectors.
Scalar triple product is found by finding the determinant of three vectors,
Hence the scalar triple product of three vectors is $\left| {\begin{array}{*{20}{c}}{ - 2}&2&0\\0&2&0\\1&{ - 2}&3\end{array}} \right|$
Now let us solve the determinant to find the final answer,
$= - 2\left[ {\left( {2 \times 3} \right) - \left( {0 \times - 2} \right)} \right] - 2\left[ {\left( {0 \times 3} \right) - \left( {0 \times 1} \right)} \right] + 0\left[ {\left( {0 \times - 2} \right) - \left( {2 \times 1} \right)} \right]$
The scalar triple product of given vector is $= - 12$
We know that 3 vectors are coplanar, if their scalar triple product is zero.
But here the scalar triple product $\ne 0$
So the condition of coplanarity fails.
Hence, we can say that the given points are not coplanar to each other.

Additional Information: Coplanar vectors are the vectors which lie on the same plane, in a three-dimensional space. These are vectors which are parallel to the same plane. We can always find in a plane any two random vectors, which are coplanar.

Note: Here while finding the determinant value we take first row and solve the problem we can also solve it by taking any other rows and columns. But the answer of the determinant never changes.