You may already be familiar with this question type (solving equations with variables in the exponents), but there's something a little different about this question.
With most other questions, the base is a constant (i.e., NOT a variable).
For example: $5^{2x+1} = 5^{x-3}$
Here, we can automatically conclude that $2x+1 = x-3$, which we can then solve for $x$
This conclusion is based on the following property: If $b^x = b^y$, then $x = y$
IMPORTANT: The above property has 3 provisos: $b \neq 0, b \neq 1$ and $b \neq -1$

These provisos should make sense.
For example, $1^x$ always equals 1. So, if an equation has $1$ in the base, we really can't make any conclusions about the exponents.
So, if $1^x = 1^3$, we can't then conclude that $x=3$, since ANY value of $x$ will satisfy the equation:
In other words, since $1^x = 1^3=1^8=1^2=1^1=1^{-2} = 1...$ etc, we can't make any conclusions about the value of $x$

This same principle also applies to equations in which the base is $0$ and $-1$.

Given all of this, we need to consider a few different cases:

case i: $x$ does not equal $0$, $1$ or $-1$
This means we can safely apply the above property to conclude that: $3x^2-5x-7=2x^2-3x+1$
Subtract $2x^2$ from both sides of the equation to get: $x^2-5x-7=-3x+1$
Add $3x$ to both sides of the equation to get: $x^2-2x-7=1$
Subtract $1$ from both sides of the equation to get: $x^2-2x-8=0$
Factor: $(x-4)(x+2)=0$
Solve: $x=4$ and $x=-2$

case ii: $x = 0$
Since the base is one of the forbidden values ($0$), we can't apply the useful property above.
Instead, we can plug $x=0$ into the original equation to determine whether it's a solution.
When we do this we get: $0^{3(0^2)-5(0)-7}=0^{2(0^2)-3(0)+1}$
When we simplify, we get: $0^{-7}=0^{1}$
Evaluate to get: $\frac{1}{0}=0$.
Since $\frac{1}{0}$ is undefined, $x = 0$ is not a solution to the given equation.

Aside: Since the question asks us to find the SUM of all possible solutions, we didn't really need to check whether $x=0$ is a solution, since it won't have any effect on the SUM. I just wanted to demonstrate the steps necessary to determining ALL possible solutions

case iii: $x = 1$
Once again, to determine whether $x=1$ is a possible solution, we'll plug $x=1$ into the original equation.
We get: $1^{3(1^2)-5(1)-7}=1^{2(1^2)-3(1)+1}$
Simplify: $1^{-9}=1^{0}$
Evaluate to get: $1=1$. WORKS!
So, $x=1$ is another solution

Aside: We don’t need to actually test $x = 1$, since 1 raised to any power will always be 1. I’m just showing all possible steps.

case iv: $x = -1$
Once again, to determine whether $x=-1$ is a possible solution, we'll plug $x=-1$ into the original equation.
We get: $(-1)^{3(-1)^2)-5(-1)-7}=(-1)^{2(-1)^2)-3(-1)+1}$
Simplify: $(-1)^{1}=(-1)^{6}$
Evaluate to get: $-1=1$. DOESN'T WORK
So, $x=-1$ is NOT a solution

The sum of all possible solutions = $4 + (-2) + 1 = 3$

Answer: C

Cheers,
Brent