$$\begin{gathered} \text{Let }f\mathrm{(x)}=x^3+x^2{f}^{\prime }\mathrm{(1)}+2x{f}^{″}(2)+f^{‴}(3),x\in R. \\ \text{Then the value of}{f}^{\prime }\mathrm{(5) }\text{is} \end{gathered}$$

The correct answer is \frac{117}{5}.

Step-by-step Explanation:

Given the function:
f\left(x\right)=x^3+x^2{f}^{\prime }\left(1\right)+2x{f}^{″}\left(2\right)+f^{‴}\left(3\right)

Notice that f^{\prime }\left(1\right), f^{″}\left(2\right), and f^{‴}\left(3\right) are constant values. Let us define them as:
a=f^{\prime }\left(1\right)
b=f^{″}\left(2\right)
c=f^{‴}\left(3\right)

Rewriting the function f\left(x\right) using these constants:
f\left(x\right)=x^3+a{x}^2+2bx+c

Now, let us find the first, second, and third derivatives of f\left(x\right) with respect to x:

1. First Derivative:
f^{\prime }\left(x\right)=3x^2+2ax+2b

2. Second Derivative:
f^{″}\left(x\right)=6x+2a

3. Third Derivative:
f^{‴}\left(x\right)=6

Now, substitute the values of x=3, x=2, and x=1 into their respective derivatives to find the constants:

From the third derivative:
c=f^{‴}\left(3\right)=6

From the second derivative at x=2:
b=f^{″}\left(2\right)=6\left(2\right)+2a=12+2a

From the first derivative at x=1:

a=3+2a+2b
a+2b+3=0

Substitute b=12+2a into the equation a+2b+3=0:
a+2(12+2a)+3=0
a+24+4a+3=0
5a+27=0
a=-\frac{27}{5}

Now, determine the value of b:
b=12+2\left(-\frac{27}{5}\right)=12-\frac{54}{5}=\frac{6}{5}

Finally, we calculate f^{\prime }\left(5\right):
f^{\prime }\left(5\right)=3\left(5^2\right)+2a\left(5\right)+2b
f^{\prime }\left(5\right)=75+10a+2b
f^{\prime }\left(5\right)=75+10\left(-\frac{27}{5}\right)+2\left(\frac{6}{5}\right)
f^{\prime }\left(5\right)=75-54+\frac{12}{5}
f^{\prime }\left(5\right)=21+\frac{12}{5}=\frac{105+12}{5}=\frac{117}{5}

Thus, the value of f^{\prime }\left(5\right) is equal to \frac{117}{5}.