$\text{If}\overrightarrow{\mathrm{ a }}\text{and}\overrightarrow{\mathrm{ b}}\text{ are two vectors such that}\left|\overrightarrow{a}\right|=7\text{ and}\left|\overrightarrow{b}\right|=4,\text{then the value of scalar product of vectors}\mathrm{ 2}\overrightarrow{a}-3\overrightarrow{b}\text{ and}\mathrm{ 2}\overrightarrow{a}+3\overrightarrow{b}\text{is}$

The correct option is 52.

To find the scalar product (dot product) of the vectors $2\overrightarrow{a}-3\overrightarrow{b}$ and $2\overrightarrow{a}+3\overrightarrow{b}$, we can use the distributive property of the scalar product (dot product), which is similar to the algebraic identity $(x-y)(x+y)=x^2-y^2$.

Let's write down the expression for the scalar product:
$$(2\overrightarrow{a}-3\overrightarrow{b})\cdot (2\overrightarrow{a}+3\overrightarrow{b})$$

Expanding the dot product using the distributive property, we get:
$$\left(2\overrightarrow{a}\right)\cdot \left(2\overrightarrow{a}\right)+\left(2\overrightarrow{a}\right)\cdot \left(3\overrightarrow{b}\right)-\left(3\overrightarrow{b}\right)\cdot \left(2\overrightarrow{a}\right)-\left(3\overrightarrow{b}\right)\cdot \left(3\overrightarrow{b}\right)$$

Using the properties of the scalar product, specifically that $\overrightarrow{u}\cdot \overrightarrow{v}=\overrightarrow{v}\cdot \overrightarrow{u}$ (commutativity) and $\overrightarrow{u}\cdot \overrightarrow{u}={\left|\overrightarrow{u}\right|}^2$, we can simplify the expression:
$$4(\overrightarrow{a}\cdot \overrightarrow{a})+6(\overrightarrow{a}\cdot \overrightarrow{b})-6(\overrightarrow{b}\cdot \overrightarrow{a})-9(\overrightarrow{b}\cdot \overrightarrow{b})$$

Since $\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{b}\cdot \overrightarrow{a}$, the middle terms cancel out:
$$4{\left|\overrightarrow{a}\right|}^2-9{\left|\overrightarrow{b}\right|}^2$$

We are given that $\left|\overrightarrow{a}\right|=7$ and $\left|\overrightarrow{b}\right|=4$. Now, substitute these values into the simplified expression:
$$4{\left(7\right)}^2-9{\left(4\right)}^2$$

Calculate the squares:
$$4\left(49\right)-9\left(16\right)$$

Multiply the terms:
$$196-144$$

Perform the subtraction:
$$196-144=52$$

Therefore, the value of the scalar product is 52.