If the points \[A\left( 1,-2,2 \right),B\left( 3,1,1 \right),C\left( -1,p,3 \right)\] are collinear, then find the value of p.

Hint: We start solving the problem by recalling the condition of collinearity of three points $\left( a,b,c \right)$, $\left( d,e,f \right)$ and $\left( g,h,i \right)$ as $\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right|=0$. We then substitute the coordinates of the given points in this determinant and then expand it. We then make the necessary calculations to get the required value of p.

Complete step-by-step answer:
According to the problem, we are given the points \[A\left( 1,-2,2 \right)\], \[B\left( 3,1,1 \right)\], \[C\left( -1,p,3 \right)\].
As we already know that the points which lie on the same line are called collinear points and the condition of collinearity for three points $\left( a,b,c \right)$, $\left( d,e,f \right)$ and $\left( g,h,i \right)$ is $\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right|=0$.

So, let us substitute the coordinates of the points \[A\left( 1,-2,2 \right)\], \[B\left( 3,1,1 \right)\], \[C\left( -1,p,3 \right)\] in the determinant.
So, we get $\left| \begin{matrix}
   1 & -2 & 2 \\
   3 & 1 & 1 \\
   -1 & p & 3 \\
\end{matrix} \right|=0$.
$\Rightarrow 1\times \left| \begin{matrix}
   1 & 1 \\
   p & 3 \\
\end{matrix} \right|-\left( -2 \right)\times \left| \begin{matrix}
   3 & 1 \\
   -1 & 3 \\
\end{matrix} \right|+2\times \left| \begin{matrix}
   3 & 1 \\
   -1 & p \\
\end{matrix} \right|=0$.
$\Rightarrow 1\times \left( \left( 1\times 3 \right)-\left( 1\times p \right) \right)-\left( -2 \right)\times \left( \left( 3\times 3 \right)-\left( -1\times 1 \right) \right)+2\times \left( \left( 3\times p \right)-\left( -1\times 1 \right) \right)=0$
$\Rightarrow \left( 3-p \right)+\left( 2 \right)\times \left( \left( 9 \right)-\left( -1 \right) \right)+2\times \left( \left( 3p \right)-\left( -1 \right) \right)=0$.
$\Rightarrow 3-p+\left( 2 \right)\times \left( 10 \right)+2\times \left( 3p+1 \right)=0$.
$\Rightarrow 3-p+20+6p+2=0$.
$\Rightarrow 5p=-25$.
$\Rightarrow p=\dfrac{-25}{5}$.
$\Rightarrow p=-5$.
We have found the value of p as 5. Hence, The value of p as 5.

Note: This is one of the ways to solve for the value of p if the given points are collinear. We can also solve this problem by using the fact that the D.R’s (Direction Ratios) of the line segment AB is multiple of D.R’s (Direction Ratios) of the line segment BC as shown below.
We know that D.R’s (Direction Ratios) of the line joining points $\left( a,b,c \right)$ and $\left( d,e,f \right)$ is $\left( d-a,e-b,f-c \right)$.
So, the D.R’s (Direction Ratios) of the line segment AB = $\left( 3-1,1+2,1-2 \right)=\left( 2,3,-1 \right)$.
Now, the D.R’s (Direction Ratios) of the line segment BC = $\left( -1-3,p-1,3-1 \right)=\left( -4,p-1,2 \right)$
Now, we have D.R’s of the line segment AB = $\lambda \times $ (D.R’s of the line segment BC).
So, we have $\left( 2,3,-1 \right)=\lambda \times \left( -4,p-1,2 \right)$.
$\Rightarrow \left( 2,3,-1 \right)=\left( -4\lambda ,p\lambda -\lambda ,2\lambda \right)$.
So, we have $-4\lambda =2$.
$\lambda =\dfrac{2}{-4}=\dfrac{-1}{2}$.
Now, we have $p\left( \dfrac{-1}{2} \right)-\left( \dfrac{-1}{2} \right)=3$.
$\Rightarrow \dfrac{-p+1}{2}=3$.
$\Rightarrow -p+1=6$.
$\Rightarrow p=-5$.