Quadratic Programming with Linear Equality Constraints
Let's solve a more general problem (Least Squares with Linear Equality Constraints):

$$
\begin{alignat*}{3}
\arg \min_{x} &amp; \quad &amp; \frac{1}{2} \left\| A x - b \right\|_{2}^{2} \\
\text{subject to} &amp; \quad &amp; C x = d
\end{alignat*}
$$

The Lagrangian is given by:

$$ L \left( x, \nu \right) = \frac{1}{2} \left\| A x - b \right\|_{2}^{2} + {\nu}^{T} \left( C x - d \right) $$

From KKT Conditions the optimal values of $ \hat{x}, \hat{\nu} $ obeys:

$$ \begin{bmatrix}
{A}^{T} A &amp; {C}^{T} \\ 
C &amp; 0
\end{bmatrix} \begin{bmatrix}
\hat{x} \\ 
\hat{\nu}
\end{bmatrix} = \begin{bmatrix}
{A}^{T} b \\ 
d
\end{bmatrix} $$

The trick here is to have a look at:

$$ \frac{1}{2} \left\| A x - b \right\|_{2}^{2} = \frac{1}{2} {x}^{T} {A}^{T} A x - {x}^{T} A b + \frac{1}{2} {b}^{T} b $$

So if we set $ w = x $, $ b = \boldsymbol{0} $, $ X = C $, $ d = \boldsymbol{1} $ and $ \frac{1}{2} {A}^{T} A = R $ then your solution is given by:

$$ \begin{bmatrix}
2 R &amp; {X}^{T} \\ 
X &amp; 0
\end{bmatrix} \begin{bmatrix}
\hat{w} \\ 
\hat{\nu}
\end{bmatrix} = \begin{bmatrix}
\boldsymbol{0} \\ 
\boldsymbol{1}
\end{bmatrix} $$

This is easy to solve in MATLAB or Python.

Handling Large System

As a request from @user5045 I add some info how to handle this in case $ R $ and $ X $ are large matrices.

We basically need to solve large scale matrix equation:

$$ \begin{bmatrix}
2 R &amp; {X}^{T} \\ 
X &amp; 0
\end{bmatrix} \begin{bmatrix}
\hat{w} \\ 
\hat{\nu}
\end{bmatrix} = \begin{bmatrix}
\boldsymbol{0} \\ 
\boldsymbol{1}
\end{bmatrix} = e = F g $$

The way to solve it is using an iterative solver.
I case $ R $ is a PSD matrix then the solver should be Preconditioned Conjugate Gradient. In MATLAB it is implemented using pcg().
In case $ R $ is only symmetric one should use Minimum Residual Solver. In MATLAB it is implemented as minres(). For a a general square matrix I'd go with cgs().