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**Hint:**To find the value of $\left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|$, we have to convert it in the form of \[\dfrac{{{z}_{2}}}{{{z}_{1}}}\] using appropriate operations, because we know the values of \[\dfrac{{{z}_{2}}}{{{z}_{1}}}\]. Also, we have to make use of the formula |z|= \[\sqrt{{{a}^{2}}+{{b}^{2}}}\] where |z| is called modulus of \[z=a+ib\]

**Complete step by step answer:**

The question demands that, we have to find the value of the term $\left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|$. Let the value of this term be ‘y’. Therefore, we will get,

\[y=\left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|.............(i)\]

We are also given in question that $\dfrac{5{{z}_{2}}}{7{{z}_{1}}}$ is purely imaginary. This means we can say that we can represent $\dfrac{5{{z}_{2}}}{7{{z}_{1}}}$ solely in terms of I (iota). Thus,

$\dfrac{5{{z}_{2}}}{7{{z}_{1}}}$= \[ki\]

Where, k is any real number. We can also write the above equation as:

\[\dfrac{{{z}_{2}}}{{{z}_{1}}}=\dfrac{7ki}{5}...............(ii)\]

Now we come back to equation (i). Now we will divide both the numerator and denominator by ‘z’. After doing this we get: -

\[y=\left| \begin{align}

& \dfrac{2{{z}_{1}}+3{{z}_{x}}}{{{z}_{1}}} \\

& \overline{\,\,\dfrac{2{{z}_{1}}+3{{z}_{z}}}{{{z}_{1}}}} \\

\end{align} \right|\]

\[\Rightarrow y=\left| \dfrac{2+3\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}{2-3\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)} \right|..................(iii)\]

Now, we will substitute value of \[\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)\] from equation (ii) into equation (iii). After doing this we will get:

$\Rightarrow y=\left| \dfrac{2+3\left( \dfrac{7{{k}_{1}}}{5} \right)}{2-3\left( \dfrac{7{{k}_{1}}}{5} \right)} \right|$

On simplifying we will get: -

$\Rightarrow y=\left| \dfrac{2+\dfrac{21ki}{5}}{2-\dfrac{21ki}{5}} \right|$

On taking Lcm and cancelling 5 from both numerator and denominator, we get: -

$\Rightarrow y=\left| \dfrac{10+21ki}{10-21ki} \right|.............(iv)$

Here, we are going to use a property of modulus which is shown below:

$\dfrac{|{{z}_{1}}|}{|{{z}_{2}}|}=\dfrac{|{{z}_{1}}|}{|{{z}_{2}}|}$

In our case ${{z}_{1}}=10+21ki$ and ${{z}_{2}}=10-21ki$

After using this identity, we will get:

$\Rightarrow y=\dfrac{|10+21ki|}{|10-21ki|}....................(v)$

In the above equation, we have to find the modulus of two terms in numerator and denominator respectively. If a complex number is z=a+ib then its modulus is given by the formula:

$|z|=\sqrt{{{a}^{2}}+{{b}^{2}}}$

Therefore, using above formula, we get: -

\[\begin{align}

& \Rightarrow y=\dfrac{\sqrt{{{\left( 10 \right)}^{2}}+{{\left( 21k \right)}^{2}}}}{\sqrt{{{\left( 10 \right)}^{2}}+{{\left( -21k \right)}^{2}}}} \\

& \Rightarrow y=\dfrac{\sqrt{100+441{{k}^{2}}}}{\sqrt{100+441{{k}^{2}}}} \\

& \Rightarrow y=1 \\

& \Rightarrow \left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right| \\

\end{align}\]

**So, the correct answer is “Option d”.**

**Note:**We cannot use the modulus formula directly in the starting as shown below: -

$\left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|=\dfrac{\sqrt{{{\left( z \right)}^{2}}+{{\left( 3 \right)}^{2}}}}{\sqrt{{{\left( a \right)}^{2}}+{{\left( -3 \right)}^{2}}}}=\dfrac{\sqrt{13}}{\sqrt{13}}=1$

In this case, the answer is the same but the method is wrong because ${{z}_{1}}$ and ${{z}_{2}}$ are both complex numbers. We will have to convert the numerator and denominator into a single complex number of the form a+ib then only we can apply the modulus formula.

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