Find the value of the following expression:
$$\left[18-3\times \left\{3-3\times \left(3÷\frac{1}{3}-3\right)+3÷\frac{1}{3}+3\right\}-4\times 2\right]$$

To solve the expression, we apply the BODMAS rule (Brackets, Of, Division, Multiplication, Addition, Subtraction):
Given expression:

$$\left[18-3\times \left\{3-3\times \left(3÷\frac{1}{3}-3\right)+3÷\frac{1}{3}+3\right\}-4\times 2\right]$$

Step 1: Solve the innermost parenthesis ()

$$\left(3÷\frac{1}{3}-3\right)=(3\times 3-3)=9-3=6$$

Substitute this back into the expression:

$$\left[18-3\times \left\{3-3\times 6+3÷\frac{1}{3}+3\right\}-4\times 2\right]$$

Step 2: Solve the operations inside the curly braces {}
Perform division and multiplication first:

$$3\times 6=18$$

$$3÷\frac{1}{3}=3\times 3=9$$

Now rewrite the curly braces:

$$\left\{3-18+9+3\right\}=\{15-18\}=-3$$

Substitute this back into the expression:

$$\left[18-3\times (-3)-4\times 2\right]$$

Step 3: Solve the remaining operations inside the square bracket []
Perform the multiplications:

$$-3\times (-3)=+9$$

$$-4\times 2=-8$$

Now substitute and solve:

$$\left[18+9-8\right]=[27-8]=19$$