Show that the four points A, B, C and D with position vectors $4\widehat{i}+5\widehat{j}+\widehat{k},-\widehat{j}-\widehat{k},3\widehat{i}+9\widehat{j}+4\widehat{k}$ and $4\left( -\widehat{i}+\widehat{j}+\widehat{k} \right)$ respectively are coplanar.

Hint: First, before proceeding for this, we must know the following formula to calculate distance between the two vectors which are $X\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)$ and $Y\left( d\widehat{i}+e\widehat{j}+f\widehat{k} \right)$is given by $\overrightarrow{XY}=\left( d-a \right)\widehat{i}+\left( e-b \right)\widehat{j}+\left( f-c \right)\widehat{k}$. Then, by using the same formula for the three distances, we want to calculate $\overrightarrow{AB},\overrightarrow{AC}$and $\overrightarrow{AD}$. Then, we know from the concept of vectors that if the determinant of directional vector from a single point is zero then all of the vectors are coplanar and by using them we prove the required statement.

Complete step-by-step answer:
In this question, we are supposed to prove that four points A, B, C and D with position vectors $4\widehat{i}+5\widehat{j}+\widehat{k},-\widehat{j}-\widehat{k},3\widehat{i}+9\widehat{j}+4\widehat{k}$ and $4\left( -\widehat{i}+\widehat{j}+\widehat{k} \right)$ respectively are coplanar.
So, before proceeding for this, we must know the following formula to calculate distance between the two vectors which are $X\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)$ and $Y\left( d\widehat{i}+e\widehat{j}+f\widehat{k} \right)$is given by:
$\overrightarrow{XY}=\left( d-a \right)\widehat{i}+\left( e-b \right)\widehat{j}+\left( f-c \right)\widehat{k}$
Now, by using the same formula for the three distances, we want to calculate as $\overrightarrow{AB},\overrightarrow{AC}$and $\overrightarrow{AD}$as:
$\begin{align}
  & \overrightarrow{AB}=\left( 0-4 \right)\widehat{i}+\left( -1-5 \right)\widehat{j}+\left( -1-1 \right)\widehat{k} \\
 & \Rightarrow \overrightarrow{AB}=-4\widehat{i}-6\widehat{j}-2\widehat{k} \\
 & \overrightarrow{AC}=\left( 3-4 \right)\widehat{i}+\left( 9-5 \right)\widehat{j}+\left( 4-1 \right)\widehat{k} \\
 & \Rightarrow \overrightarrow{AC}=-\widehat{i}+4\widehat{j}+3\widehat{k} \\
 & \overrightarrow{AD}=\left( -4-4 \right)\widehat{i}+\left( 4-5 \right)\widehat{j}+\left( 4-1 \right)\widehat{k} \\
 & \Rightarrow \overrightarrow{AD}=-8\widehat{i}-\widehat{j}+3\widehat{k} \\
\end{align}$
Now, we know from the concept of vectors that if the determinant of directional vector from a single point is zero then all of the vectors are coplanar.
So, now we will find the determinant of the vectors found above by using the concept of determinants as:
$\left| \begin{matrix}
   -4 & -6 & -2 \\
   -1 & 4 & 3 \\
   -8 & -1 & 3 \\
\end{matrix} \right|$
Now, we will calculate the value of the above determinant to check whether the given vectors are coplanar or not.
So, by expanding with first row to calculate the determinant, we get:
$-4\left( 4\times 3-3\times \left( -1 \right) \right)+6\left( 3\times \left( -1 \right)-3\times \left( -8 \right) \right)-2\left( \left( -1 \right)\times \left( -1 \right)-4\times \left( -8 \right) \right)$
Then, by solving the above expression to get the value of determinant as:
$\begin{align}
  & -4\left( 12+3 \right)+6\left( -3+24 \right)-2\left( 1+32 \right) \\
 & \Rightarrow -4\left( 15 \right)+6\left( 21 \right)-2\left( 33 \right) \\
 & \Rightarrow -60+126-66 \\
 & \Rightarrow 0 \\
\end{align}$
So, the determinant of the above vectors is zero which proves that the given vectors are coplanar.
Hence, the given vectors are coplanar is proved.

Note: To solve these types of questions we need to be careful while calculating the determinants of the numbers as there is a high chance of mistake with the positive and negative sign. Moreover, there is no compulsion that we will expand the determinant along row one, we can expand along any row or column, we will get the same result.