The problem is given by:
$$\begin{align*}
\arg \min_{x} \quad & {x}^{T} B x \\
\text{subject to} \quad & C x = d
\end{align*}$$
Where $ B $ is Positive Semi Definite (PSD) matrix.
Since $ B $ is PSD one could write the problem as:
$$\begin{align*}
\arg \min \quad & {\left\| A x - b \right\|}_{2}^{2} \\
\text{subject to} \quad & C x = d
\end{align*}$$
Where $ {A}^{T} A = B $ (Since it is PSD it is guaranteed such $ A $ exists) and $ b = \boldsymbol{0} $.
Now, this a simple Least Squares problem with Linear Equality Constraints.
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:
$$ \begin{bmatrix}
{A}^{T} A & {C}^{T} \\
C & 0
\end{bmatrix} \begin{bmatrix}
\hat{x} \\
\hat{\nu}
\end{bmatrix} = \begin{bmatrix}
{A}^{T} b \\
d
\end{bmatrix} $$
Since $ B = {A}^{T} A $ and $ b = \boldsymbol{0} $ the above reduces to:
$$ \begin{bmatrix}
B & {C}^{T} \\
C & 0
\end{bmatrix} \begin{bmatrix}
\hat{x} \\
\hat{\nu}
\end{bmatrix} = \begin{bmatrix}
\boldsymbol{0} \\
d
\end{bmatrix} $$
Now all needed is to solve the above with any Linear System Solver.
If you want to use iterative procedure you may use Linear Least Squares with Linear Equality Constraints - Iterative Solver (With included MATLAB Code).
Best Answer
Dual Problem Solution
The Lagrangian is given by:
$$ L \left( x, \lambda, \nu \right) = {x}^{T} x + {\lambda}^{T} \left( x - a \right) + \nu \left( \boldsymbol{1}^{T} x - b \right) = {x}^{T} x + \left( \lambda + \nu \boldsymbol{1} \right)^{T} x -{\lambda}^{T} a - \nu b $$
The Dual Function is given by:
$$ g \left( \lambda, \nu \right) = \inf_{x} L \left( x, \lambda, \nu \right) $$
Looking at the term related to $ x $:
$$ \inf_{x} {x}^{T} x + \left( \lambda + \nu \boldsymbol{1} \right)^{T} x $$
Which is a quadratic form of $ x $ with its minimizer given by: $$ {x}^{\ast} = -\frac{1}{2} \left( \lambda + \nu \boldsymbol{1} \right) $$
Its minimum given by $$ {{x}^{\ast}}^{T} {x}^{\ast} + \left( \lambda + \nu \boldsymbol{1} \right)^{T} {x}^{\ast} = -\frac{1}{4} \left( \lambda + \nu \boldsymbol{1} \right)^{T} \left( \lambda + \nu \boldsymbol{1} \right) $$
Hence the Dual Problem is given by:
\begin{align*} \text{maximize} & \quad & -\frac{1}{4} \left( \lambda + \nu \boldsymbol{1} \right)^{T} \left( \lambda + \nu \boldsymbol{1} \right) - {\lambda}^{T} a - \nu b \\ \text{subject to} & \quad & \lambda \succeq 0 \end{align*}
The problem is Concave in $ \left( \lambda, \nu \right) $ hence it is a convex problem.
It can be solved by a Quadratic Programming as:
$$ \left( \lambda + \nu \boldsymbol{1} \right)^{T} \left( \lambda + \nu \boldsymbol{1} \right) = {\left\| E v \right\|}_{2}^{2} = {v}^{T} {E}^{T} E v = {v}^{T} H v $$
Where $ v = {\left[ \lambda, \nu \right]}^{T}, \; E = \left[ I, \boldsymbol{1} \right], \; H = {E}^{T} E $. Then the problem becomes:
$$ \begin{align*} \text{minimize} & \quad & \frac{1}{4} {v}^{T} H v + {v}^{T} f \\ \text{subject to} & \quad & A v \preceq 0 \end{align*} $$
Where $ A = - \left[ I, \boldsymbol{0} \right], \; f = {\left[ a, b \right]}^{T} $.
The above can directly solved by MATLAB's
quadprog(). Then $ {x}^{\ast} = -0.5 E v $.