Is there an efficient way to solve this problem? For the problem below, the question hint suggests solving for a, b, and c. I have tried to use a matrix to solve for a, b, c through row reduced echelon form but this requires alot of very tedious algebra and I was wondering if there was a more elegant solution or more elegant way of solving this problem?

 A: An alternative solution, easily generalizable to higher degree polynomials, would go like this:
First, consider the polynomial:
$P_1(x) = \frac{q_1(x - p_2)(x - p_3)}{(p_1 - p_2)(p_1 - p_3)}$.
Then, $P_1(p_1) = q_1$, while $P_1(p_2) = P_1(p_3) = 0$. By permuting the indices, you get polynomials $P_2$ and $P_3$ that equal $q_2$ and $q_3$ at $p_2$ and $p_3$ respectively, but are $0$ at the other critical points.
Then, $P = P_1 + P_2 + P_3$ has the desired properties, and solving for $a$, $b$, and $c$ is simply a (tedious) matter of expanding the polynomials and gathering like terms. To do this for more points, using a higher degree polynomial, you need to create more polynomials $P_i$ with more factors in the numerator and denominator.
A: I see what you mean.  Just to see what the result of the row reduction is I entered the relevant matrix into a computer algebra system (by typing
row reduce {{1,p_1,(p_1)^2,q_1},{1,p_2,(p_2)^2,q_2},{1,p_3,(p_3)^2,q_3}}
into Wolfram Alpha).  Here is the result.

It is interesting that the numerators of $a, b$, and $c$ is each the sum of six terms of degree 4, 3, and 2, respectively.  These numerators are kind of reminiscent of the complete homogeneous symmetric polynomials, although that is not exactly what they are.
It would be interesting to know whether there is a pattern here that would give the formula for the coefficients of the $n^{th}$ degree polynomial that goes through $(p_1, q_1), \ldots, (p_n,q_n)$ when $p_1 < p_2 < \ldots < p_n$.
That said, my short answer to your original question is, "No, I do not see a slick way to solve for $a, b$, and $c$ explicitly."  Brute force and determination is how I would approach finding these formulas for $a, b$, and $c$ if one needed them.
A: Something that seems overlooked in the present answer and comments is that the question only asked to show that such a curve exists. Nothing was said about getting an explicit formula for any such curve. Incidentally, the "Hint" is somewhat misleading. The hint should said something like "Show that it is possible to solve for $a,$ $b,$ and $c.$
Plugging the three points in gives
$$ (1): \;\;\; q_1 \; = \; a + bp_1 + cp_1^2    $$
$$ (2): \;\;\; q_2 \; = \; a + bp_2 + cp_2^2    $$
$$ (3): \;\;\; q_3 \; = \; a + bp_3 + cp_3^2    $$
Subtracting equations as indicated gives
$$ (2) - (1): \;\;\;\; q_2 - q_1 \; = \; b(p_2 - p_1) + c(p_2^2 - p_1^2)      $$
$$ (3) - (2): \;\;\;\; q_3 - q_2 \; = \; b(p_3 - p_2) + c(p_3^2 - p_2^2)      $$
Now divide both sides of $\;(2) - (1)$ by $p_2 - p_1,\;$ and divide both sides of $\;(3) - (2)$ by $\;p_3 - p_2,\;$ to get
$$ (4): \;\;\; \frac{q_2 - q_1}{p_2 - p_1} \; = \; b + c(p_2 + p_1)    $$
$$ (5): \;\;\; \frac{q_3 - q_2}{p_3 - p_2} \; = \; b + c(p_3 + p_2)    $$
Now subtract equations (4) and (5) to get
$$ (5) - (4): \;\;\;\; \frac{q_3 - q_2}{p_3 - p_2} \; - \; \frac{q_2 - q_1}{p_2 - p_1} \;\; = \;\; c(p_3 - p_1) $$
At this point it is easy to show that there exists a solution for $a,\,b,\, c$ in terms of the coordinates of the points, and thus there exists such a curve. In fact, there exists exactly one such solution (more than what we were asked to show), and thus there exists exactly one such (quadratic) curve.
More specifically, since none of the $p_k$'s are equal to each other, none of the denominators in sight are zero. From equation $\;(5) - (4),$ it is clear that we can solve for $c.$ With $c$ known, either of equations $(4)$ or $(5)$ can be solved for $b$ by subtracting a single expression from both sides (the expression that has $c$ in it). Note that in carrying this out we don't need to worry about anything being undefined (e.g. dividing by an expression that could be zero, or doing something else not always defined). Finally, with $b$ and $c$ known, any of equations $(1),\,(2),\,(3)$ can be solved for $a$ by subtracting two expressions from both sides (and again, we don't have to worry about anything being undefined).
