The correct option is 24.

Analysis of the Given Problem:
From the provided image, we are given the following vectors and relations:
1. Vector $\overrightarrow{a}=\hat{i}+2\hat{j}+\hat{k}$
2. Vector $\overrightarrow{b}=3(\hat{i}-\hat{j}+\hat{k})$
3. Dot product $\overrightarrow{a}\cdot \overrightarrow{c}=3$
4. Cross product relationship $\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{b}$

We need to evaluate the scalar expression:
$\overrightarrow{a}\cdot \left[\left(\overrightarrow{c}\times \overrightarrow{b}\right)-\overrightarrow{b}-\overrightarrow{c}\right]$

Step 1: Expand the expression using the distributive property of the dot product
Using vector algebra properties, we can distribute the dot product with $\overrightarrow{a}$ across the terms inside the square brackets:
$\overrightarrow{a}\cdot \left[\left(\overrightarrow{c}\times \overrightarrow{b}\right)-\overrightarrow{b}-\overrightarrow{c}\right]=\overrightarrow{a}\cdot \left(\overrightarrow{c}\times \overrightarrow{b}\right)-\overrightarrow{a}\cdot \overrightarrow{b}-\overrightarrow{a}\cdot \overrightarrow{c}$   — (Equation 1)

Step 2: Evaluate the term $\overrightarrow{a}\cdot \overrightarrow{b}$
We are given that $\overrightarrow{b}=\overrightarrow{a}\times \overrightarrow{c}$.
Since the cross product of any two vectors is perpendicular to both individual vectors, $\overrightarrow{b}$ must be perpendicular to $\overrightarrow{a}$.
Therefore, the dot product of $\overrightarrow{a}$ and $\overrightarrow{b}$ is zero:
$\overrightarrow{a}\cdot \overrightarrow{b}=0$

Step 3: Evaluate the scalar triple product term $\overrightarrow{a}\cdot \left(\overrightarrow{c}\times \overrightarrow{b}\right)$
By the cyclic property of the scalar triple product:
$\overrightarrow{a}\cdot \left(\overrightarrow{c}\times \overrightarrow{b}\right)=\left(\overrightarrow{a}\times \overrightarrow{c}\right)\cdot \overrightarrow{b}$
Substituting $\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{b}$ into the equation:
$\left(\overrightarrow{a}\times \overrightarrow{c}\right)\cdot \overrightarrow{b}=\overrightarrow{b}\cdot \overrightarrow{b}={\left|\overrightarrow{b}\right|}^2$

Now, let us calculate the magnitude squared of vector $\overrightarrow{b}$:
$\overrightarrow{b}=3\hat{i}-3\hat{j}+3\hat{k}$
${\left|\overrightarrow{b}\right|}^2=3^2+{\left(-3\right)}^2+3^2=9+9+9=27$
Thus, we have:
$\overrightarrow{a}\cdot \left(\overrightarrow{c}\times \overrightarrow{b}\right)=27$

Step 4: Substitute the values back into Equation 1
Using the values obtained:
1. $\overrightarrow{a}\cdot \left(\overrightarrow{c}\times \overrightarrow{b}\right)=27$
2. $\overrightarrow{a}\cdot \overrightarrow{b}=0$
3. $\overrightarrow{a}\cdot \overrightarrow{c}=3$ (given)

We get:
$\overrightarrow{a}\cdot \left[\left(\overrightarrow{c}\times \overrightarrow{b}\right)-\overrightarrow{b}-\overrightarrow{c}\right]=27-0-3=24$