Answer: D. $x^2 - 22x + 121 = 0$

Explanation:1. Understand the problem: A quadratic equation has exactly one distinct real solution if its discriminant ($\Delta = b^2 - 4ac$) is equal to zero. Alternatively, this occurs if the quadratic is a perfect square trinomial.

2. Evaluate the discriminant for each given quadratic equation ($ax^2+bx+c=0$): [A] $x^2 - 22 = 0$ Here, $a=1$, $b=0$, $c=-22$. $\Delta = (0)^2 - 4(1)(-22) = 0 + 88 = 88$. Since $\Delta > 0$, this equation has two distinct real solutions.

[B] $x^2 - 11x = 0$ Here, $a=1$, $b=-11$, $c=0$. $\Delta = (-11)^2 - 4(1)(0) = 121 - 0 = 121$. Since $\Delta > 0$, this equation has two distinct real solutions (which are $x=0$ and $x=11$).

[C] $7x^2 - 11x + 121 = 0$ Here, $a=7$, $b=-11$, $c=121$. $\Delta = (-11)^2 - 4(7)(121) = 121 - 28 \times 121 = 121(1 - 28) = 121(-27)$. Since $\Delta < 0$, this equation has no real solutions.

[D] $x^2 - 22x + 121 = 0$ Here, $a=1$, $b=-22$, $c=121$. $\Delta = (-22)^2 - 4(1)(121) = 484 - 484 = 0$. Since $\Delta = 0$, this equation has exactly one distinct real solution. Alternatively, notice that $x^2 - 22x + 121$ is a perfect square trinomial: $(x - 11)^2 = 0$. The only solution is $x=11$.

3. Conclusion: The quadratic equation that has exactly one distinct real solution is $x^2 - 22x + 121 = 0$, which corresponds to option [D].