sat-dog

Choice C is correct. It's given that the graphs of the equations in the given system intersect at exactly one point, $\left(x,y\right)$, in the xy-plane. Therefore, $\left(x,y\right)$ is the only solution to the given system of equations. The given system of equations can be solved by subtracting the second equation, $y=3x+a$, from the first equation, $y=2x^2-21x+64$. This yields $y-y=\left(2x^2-21x+64\right)-\left(3x+a\right)$, or $0=2x^2-24x+64-a$. Since the given system has only one solution, this equation has only one solution. A quadratic equation in the form $r{x}^2+sx+t=0$, where $r$, $s$, and $t$ are constants, has one solution if and only if the discriminant, $s^2-4rt$, is equal to zero. Substituting $2$ for $r$, $-24$ for $s$, and $-a+64$ for $t$ in the expression $s^2-4rt$ yields ${\left(-24\right)}^2-\left(4\right)\left(2\right)\left(64-a\right)$. Setting this expression equal to zero yields ${\left(-24\right)}^2-\left(4\right)\left(2\right)\left(64-a\right)=0$, or $8a+64=0$. Subtracting $64$ from both sides of this equation yields $8a=-64$. Dividing both sides of this equation by $8$ yields $a=-8$. Substituting $-8$ for $a$ in the equation $0=2x^2-24x+64-a$ yields $0=2x^2-24x+64+8$, or $0=2x^2-24x+72$. Factoring $2$ from the right-hand side of this equation yields $0=2\left(x^2-12x+36\right)$. Dividing both sides of this equation by $2$ yields $0=x^2-12x+36$, which is equivalent to $0=\left(x-6\right)\left(x-6\right)$, or $0={\left(x-6\right)}^2$. Taking the square root of both sides of this equation yields $0=x-6$. Adding $6$ to both sides of this equation yields $x=6$. 

Choice A is incorrect. This is the value of $a$, not $x$.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.