If $$x = 1$$ is a critical point of the function $$f(x) = (3x^2 + ax - 2 - a)e^x$$, then:

We are given the function $$f(x)=\bigl(3x^{2}+ax-2-a\bigr)\,e^{x}$$ and we are told that $$x=1$$ is a critical point.

For a point to be critical we must have $$f'(1)=0$$. We therefore begin by differentiating $$f$$.

First recall the product rule: if $$u(x)$$ and $$v(x)$$ are functions of $$x$$, then $$(u\,v)'=u'\,v+u\,v'$$.

Let us set $$u(x)=3x^{2}+ax-2-a$$ and $$v(x)=e^{x}$$. Their derivatives are $$u'(x)=6x+a$$ and $$v'(x)=e^{x}$$. Applying the product rule we get

$$\begin{aligned} f'(x) &= u'(x)\,v(x)+u(x)\,v'(x) \\ &= \bigl(6x+a\bigr)e^{x}+\bigl(3x^{2}+ax-2-a\bigr)e^{x}. \end{aligned}$$

Both terms contain the common factor $$e^{x}$$, so we factor it out:

$$f'(x)=\bigl[\,6x+a+3x^{2}+ax-2-a\,\bigr]\,e^{x}.$$

Simplifying the bracket, we collect like terms:

$$6x+a+3x^{2}+ax-2-a = 3x^{2}+(6x+ax)+\bigl(a-a\bigr)-2 = 3x^{2}+x(6+a)-2.$$

Thus

$$f'(x)=\bigl(3x^{2}+(6+a)x-2\bigr)\,e^{x}.$$

Because $$e^{x}>0$$ for all real $$x$$, the equation $$f'(1)=0$$ is equivalent to setting the polynomial factor to zero at $$x=1$$:

$$3(1)^{2}+(6+a)(1)-2=0.$$

This simplifies to

$$3+6+a-2=0 \quad\Longrightarrow\quad 7+a=0 \quad\Longrightarrow\quad a=-7.$$

So the parameter is fixed as $$a=-7$$. Substituting this value back into $$f(x)$$, we get

$$f(x)=\bigl(3x^{2}-7x-2-(-7)\bigr)e^{x}=\bigl(3x^{2}-7x+5\bigr)e^{x}.$$

To find all critical points we again consider $$f'(x)$$, now with $$a=-7$$. The derivative remains

$$f'(x)=\bigl(3x^{2}-x-2\bigr)\,e^{x}.$$

Setting the bracket to zero, we solve the quadratic

$$3x^{2}-x-2=0.$$

The discriminant is $$\Delta=(-1)^{2}-4(3)(-2)=1+24=25,$$ so

$$x=\frac{1\pm\sqrt{25}}{2\cdot3}=\frac{1\pm5}{6}.$$

This yields the two critical points

$$x_{1}=1, \qquad x_{2}=-\dfrac{2}{3}.$$

Next we determine the nature (maxima or minima) of each point using the second-derivative test.

We have already written $$f'(x)=h(x)\,e^{x}$$ with $$h(x)=3x^{2}-x-2.$$ Differentiating once more, we remember that $$(h\,e^{x})'=(h'+h)\,e^{x}.$$ Therefore

$$f''(x)=\bigl(h'(x)+h(x)\bigr)\,e^{x}.$$

Compute $$h'(x)=6x-1,$$ so

$$h'(x)+h(x)=(6x-1)+(3x^{2}-x-2)=3x^{2}+5x-3.$$

Because $$e^{x}>0,$$ the sign of $$f''(x)$$ is exactly the sign of the quadratic $$3x^{2}+5x-3.$$

Evaluate it at each critical point:

For $$x=1$$:

$$3(1)^{2}+5(1)-3 = 3+5-3 = 5 > 0,$$

so $$f''(1)>0,$$ indicating a local minimum at $$x=1.$$

For $$x=-\dfrac{2}{3}$$:

$$3\left(\!-\dfrac{2}{3}\!\right)^{2}+5\left(\!-\dfrac{2}{3}\!\right)-3 =3\left(\dfrac{4}{9}\right)-\dfrac{10}{3}-3 =\dfrac{4}{3}-\dfrac{10}{3}-3 =-\dfrac{6}{3}-3 =-2-3=-5<0,$$

so $$f''\!\left(-\dfrac{2}{3}\right)<0,$$ indicating a local maximum at $$x=-\dfrac{2}{3}.$$

Therefore, $$x=1$$ is a point of local minima, while $$x=-\dfrac{2}{3}$$ is a point of local maxima.

Comparing with the options, this corresponds exactly to Option D.

Hence, the correct answer is Option D.