I have these data points.
$f_i(x)= \{10, 11, 14\}$
$x_i= \{0, 1, 3\}$
then the basis functions are.
$\pi_0 = 1 \\\pi_1=(x-x_0)=x \\\pi_2=(x-x_0)(x-x_1) =x(x-1)$
So the matrix will become.
$\begin{bmatrix} 1 & & & 10\\ 1& 1 & & 11\\ 1& 3 & 6 & 14 \end{bmatrix}$
When solved I get $10+x+\frac{1}{6}x^2$. Which does not give me the any of the answers in a multiple choice question for $x=2$.
I don't know the matrix method you are using, but from first principles, we can proceed as follows.
What is the simplest polynomial $f_0(x)$ that has value $10$ at $x = 0$? Clearly, $f_0(x) = 10$ works.
$f_0(1) = 10$ whereas we need a polynomial that has value $11$ instead. So, add $g_1(x) = (11-10)(x-0)/(1-0)$ to $f_0(x)$ to get $$f_1(x) = f_0(x) + g_1(x) = 10 + (11-10)\frac{x-0}{1-0} = 10 +x.$$ Note that $g_1(x)$ was carefully chosen to ensure that it evaluates to $0$ at $x=0$ so as not to mess up the fact that $f_0(x)$ is correctly computing the desired value at $x=0$, while at $x=1$, $g_1(x)$ provides just the missing amount $11-10 = 1$.
$f_1(3) = 13$ and so we are again short by $1$. So we add to $f_1(x)$ a polynomial $g_2(x)$ that we choose so that $g_2(0) = g_2(1) = 0$ while $g_2(3) = 1$. Clearly $x(x-1)$ satisfies the first two constraints, but it has value $3\times 2 = 6$ at $x=3$. So, $$f_2(x) = f_1(x) + (14-13)\frac{x(x-1)}{6} = 10+x + \frac{x(x-1)}{6} = 10 + \frac{5}{6}x + \frac{1}{6}x^2$$ is the polynomial we need to find.
Quick check: $f_2(0) = 10, f_2(1) = 10+\frac{5}{6}+\frac{1}{6} =11$, and $f_2(3)(x) = 10 + \frac{5}{2} + \frac{9}{6} = 14$ as it should be.
If you will look at your work, you will see that you used $x^2$ instead of your basis function $x(x-1)$ which is why your polynomial $10+x+\frac{1}{6}x^2$ is incorrect.