The sum of distinct values of $$\lambda$$ for which the system of equations:
$$(\lambda - 1)x + (3\lambda + 1)y + 2\lambda z = 0$$
$$(\lambda - 1)x + (4\lambda - 2)y + (\lambda + 3)z = 0$$
$$2x + (3\lambda + 1)y + 3(\lambda - 1)z = 0$$
Has non-zero solutions, is_______.

For a homogeneous linear system to have a non-zero (non-trivial) solution, the necessary and sufficient condition is that the determinant of its coefficient matrix must be zero. We therefore begin by writing the coefficient matrix of the three given equations:

$$A \;=\; \begin{vmatrix} \lambda-1 & 3\lambda+1 & 2\lambda\\ \lambda-1 & 4\lambda-2 & \lambda+3\\ 2 & 3\lambda+1 & 3(\lambda-1) \end{vmatrix}.$$

We need $$\det(A)=0.$$ To make the expansion simpler, we transform the matrix a little. Observe that if we replace the second row by (Row 2 - Row 1), the value of the determinant does not change (because we have only used the elementary operation “$$R_2 \leftarrow R_2-R_1$$”). Carrying this out we get

$$\det(A)= \begin{vmatrix} \lambda-1 & 3\lambda+1 & 2\lambda\\ 0 & (4\lambda-2)-(3\lambda+1) & (\lambda+3)-2\lambda\\ 2 & 3\lambda+1 & 3(\lambda-1) \end{vmatrix}.$$

Simplifying each new element of Row 2 gives

$$\det(A)= \begin{vmatrix} \lambda-1 & 3\lambda+1 & 2\lambda\\ 0 & \lambda-3 & -\lambda+3\\ 2 & 3\lambda+1 & 3(\lambda-1) \end{vmatrix}.$$

Now we expand the determinant along the first column (cofactor expansion). The formula for a $$3\times3$$ determinant expanded along the first column is

$$\det(A)=a_{11}M_{11}-a_{21}M_{21}+a_{31}M_{31},$$

where $$M_{ij}$$ is the minor of the element in row $$i$$, column $$j$$. In our matrix $$a_{21}=0$$, so its whole term vanishes immediately, and we have

$$\det(A) =(\lambda-1) \begin{vmatrix} \lambda-3 & -\lambda+3\\ 3\lambda+1 & 3(\lambda-1) \end{vmatrix} +\;2 \begin{vmatrix} 3\lambda+1 & 2\lambda\\ \lambda-3 & -\lambda+3 \end{vmatrix}.$$

We now compute each $$2\times2$$ minor. For any $$2\times2$$ matrix $$\begin{vmatrix}a&b\\c&d\end{vmatrix}$$ the determinant is $$ad-bc$$.

First minor:

$$$ \begin{vmatrix} \lambda-3 & -\lambda+3\\ 3\lambda+1 & 3(\lambda-1) \end{vmatrix} =(\lambda-3)\,3(\lambda-1)-(-\lambda+3)(3\lambda+1). $$$

Multiplying out each product, we get

$$(\lambda-3)3(\lambda-1)=3(\lambda-3)(\lambda-1)=3(\lambda^2-4\lambda+3)=3\lambda^2-12\lambda+9,$$

$$(-\lambda+3)(3\lambda+1)=-3\lambda^2+8\lambda+3.$$

Therefore the first minor equals

$$3\lambda^2-12\lambda+9-\bigl(-3\lambda^2+8\lambda+3\bigr)=6\lambda^2-20\lambda+6=2(3\lambda^2-10\lambda+3).$$

Second minor:

$$$ \begin{vmatrix} 3\lambda+1 & 2\lambda\\ \lambda-3 & -\lambda+3 \end{vmatrix} =(3\lambda+1)(-\lambda+3)-2\lambda(\lambda-3). $$$

Compute each product:

$$(3\lambda+1)(-\lambda+3)=-3\lambda^2+8\lambda+3,$$

$$2\lambda(\lambda-3)=2\lambda^2-6\lambda.$$

Hence the second minor becomes

$$-3\lambda^2+8\lambda+3-\bigl(2\lambda^2-6\lambda\bigr)=-5\lambda^2+14\lambda+3.$$

Putting these back into our expansion, we have

$$$ \det(A) =(\lambda-1)\,2(3\lambda^2-10\lambda+3)+2\bigl(-5\lambda^2+14\lambda+3\bigr). $$$

Factor out the common factor $$2$$ for convenience:

$$\det(A)=2\Bigl[(\lambda-1)(3\lambda^2-10\lambda+3)+(-5\lambda^2+14\lambda+3)\Bigr].$$

Next, expand the bracket $$(\lambda-1)(3\lambda^2-10\lambda+3)$$. We multiply term by term:

$$\lambda(3\lambda^2-10\lambda+3)=3\lambda^3-10\lambda^2+3\lambda,$$

$$-(3\lambda^2-10\lambda+3)=-3\lambda^2+10\lambda-3.$$

Adding them gives

$$3\lambda^3-13\lambda^2+13\lambda-3.$$

Now add the remaining polynomial inside the big brackets:

$$$ (3\lambda^3-13\lambda^2+13\lambda-3)+(-5\lambda^2+14\lambda+3) =3\lambda^3-18\lambda^2+27\lambda. $$$

Factor $$3\lambda$$ from this expression:

$$3\lambda(\lambda^2-6\lambda+9)=3\lambda(\lambda-3)^2.$$

Therefore

$$\det(A)=2\bigl[3\lambda(\lambda-3)^2\bigr]=6\lambda(\lambda-3)^2.$$

For non-trivial solutions we need $$\det(A)=0$$, hence

$$6\lambda(\lambda-3)^2=0.$$

Since the constant factor $$6$$ is never zero, we require

$$\lambda=0 \quad\text{or}\quad (\lambda-3)^2=0\;\Longrightarrow\;\lambda=3.$$

Thus the distinct values of $$\lambda$$ that allow non-zero solutions are $$0 \text{ and } 3.$$ Their sum is

$$0+3=3.$$

So, the answer is $$3$$.