Let A, B, C be three sets of complex number as defined below,
$A=\{z:im(z)\ge 1\}$
$B=\{z:\left| z-2-i \right|=3\}$
$C=\{z:\operatorname{Re}((1-i)z)=\sqrt{2}\}$
Let z be any point in $A\bigcap B\bigcap C$ and let $w$ be any point satisfying $\left| w-2-i \right|<3$. Then $\left| z \right|-\left| w \right|+3$ lies between
(a)-6 and 3
(b)-3 and 6
(c)-6 and 6
(d)-3 and 9

Hint: Evaluation the expression $\left| w-2-i \right|<3$ to find $\left| w\right|$ using the inequality $\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| {{z}_{1}} \right|- \left| {{z}_{2}} \right|$ if ${{z}_{1}}$ and ${{z}_{2}}$ be two complex numbers. Similarly evaluate $\left| z-2-i \right|=3$ which is the expression used to define set B to find $\left| z \right|$. Combine both these inequalities to find the value of $\left| z \right|-\left| w \right|+3$.

Complete step-by-step solution
In the question it is given that z be any point in $A\bigcap B\bigcap C$and let $w$ be any point satisfying $\Rightarrow \left| w-2-i \right| < 3.........(1)$.
We know that if ${{z}_{1}}$ and ${{z}_{2}}$ be two complex number then $\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|$ that is the modulus of the difference of two complex numbers can never be less than the difference of their moduli. Now using this identity in equation (1) we get,
 $\left| w-2-i \right|$ can be written as $\left| w-(2+i) \right|$
$3 > \left| w-(2+i) \right|\ge \left| w \right|-\left| 2+i \right|.........(2)$ using identity $\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|$ and equation (1)
Now equation (2) can be written as,
$\Rightarrow 3 > \left| w \right|-\left| 2+i \right|.......(3)$.
We will now find the modulus value of $\left| 2+i \right|$ which is equal to the square root of the sum of the squares of real and imaginary value of the complex no. So, if $z=a+ib$ be a complex number then modulus of z, is given by $\left| z\right|=\sqrt{{{a}^{2}}+{{b}^{2}}}$. Thus,
$\Rightarrow \left| 2+i \right|=\sqrt{{{(2)}^{2}}+{{1}^{2}}}$.
$\Rightarrow \left| 2+i \right|=\sqrt{4+1}$.
$\Rightarrow \left| 2+i \right|=\sqrt{5}$.
From equation (3) we have,
$\Rightarrow \left| w \right|-\left| 2+i \right| < 3$
$\Rightarrow -3+\left| 2+i \right| < \left| w \right| < 3+\left| 2+i \right|$.
Writing the value of $\left| 2+i \right|=\sqrt{5}$in the above inequality we get,
$\Rightarrow -3+\sqrt{5} < \left| w \right| < 3+\sqrt{5}$.
Now multiplying the whole inequality by -1, when we multiply an inequality by a negative sign number then the sign of inequality changes that means greater than significant changes to less than sign and vice versa, so the following expression will come by multiplying by -1.
$-(-3+\sqrt{5}) > -\left| w \right| > -(3+\sqrt{5})$ which can also be written as $-3-\sqrt{5} < -\left| w \right| < 3-\sqrt{5}........(4)$.
Similarly $\left| z-2-i \right|=3$ can be expanded using $\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|$ inequality. Thus
$\Rightarrow \left| z \right|-\left| 2+i \right|\le \left| z-2-i \right|=3$.
$\Rightarrow \left| z \right|-\left| 2+i \right| < 3$.
$\Rightarrow \left| z\right|-\sqrt{5} < 3$.
$\Rightarrow -3+\sqrt{5} < \left| z \right| < 3+\sqrt{5}............(5)$.
Combining equation (4) and (5) we get
$(-3+\sqrt{5})+(-3-\sqrt{5}) < \left| z \right|-\left| w \right| < (3+\sqrt{5})+(3-\sqrt{5)}$ which can further be simplified as
$\Rightarrow -6 < \left| z \right|-\left| w \right| < 6$ now adding 3 to the whole inequality
$\Rightarrow -6+3 < \left| z \right|-\left| w \right|+3 < 6+3$.
$\Rightarrow -3 < \left| z \right|-\left| w \right|+3 < 9$.
Hence, the value of $\left| z \right|-\left| w \right|+3$ will lie between -3 and 9. So, option (d) is correct.

Note: When we multiplying the whole inequality by -1 that is by a negative number then the sign of inequality changes that means greater than sign changes to less than sign and vice versa, so when we will multiply by -1 to the expression $-3+\sqrt{5} < \left| w \right| < 3+\sqrt{5}$ we get $-(-3+\sqrt{5}) > -\left| w \right| > -(3+\sqrt{5})$ not $-(-3+\sqrt{5}) < -\left| w \right| < -(3+\sqrt{5})$.