If the system of linear equations $$2x + 2y + 3z = a$$, $$3x - y + 5z = b$$, $$x - 3y + 2z = c$$ where $$a, b, c$$ are non-zero real numbers, has more than one solution, then

We are told that the three simultaneous linear equations

$$2x + 2y + 3z = a,$$

$$3x - y + 5z = b,$$

$$x - 3y + 2z = c,$$

possess more than one solution. For a system of three equations in three unknowns this can happen only when there are infinitely many solutions, and that in turn occurs precisely when the determinant of the coefficient matrix is zero and the augmented matrix remains consistent. We therefore start by writing the coefficient matrix

$$A=\begin{pmatrix}2 & 2 & 3\\[2pt] 3 & -1 & 5\\[2pt] 1 & -3 & 2\end{pmatrix}$$

and evaluating its determinant. Using expansion along the first row (Laplace expansion), we have

$$|A| \;=\; 2\begin{vmatrix}-1 & 5\\ -3 & 2\end{vmatrix} -\;2\begin{vmatrix}3 & 5\\ 1 & 2\end{vmatrix} +\;3\begin{vmatrix}3 & -1\\ 1 & -3\end{vmatrix}.$$

Now we compute each of the $$2\times2$$ determinants one by one.

First minor:

$$\begin{vmatrix}-1 & 5\\ -3 & 2\end{vmatrix}=(-1)(2)-(5)(-3)=-2+15=13.$$ So the first term gives $$2\times13=26.$$

Second minor:

$$\begin{vmatrix}3 & 5\\ 1 & 2\end{vmatrix}=3\cdot2-5\cdot1=6-5=1.$$ So the second term gives $$-\,2\times1=-2.$$

Third minor:

$$\begin{vmatrix}3 & -1\\ 1 & -3\end{vmatrix}=3(-3)-(-1)(1)=-9+1=-8.$$ So the third term gives $$3\times(-8)=-24.$$

Putting all three contributions together we obtain

$$|A| = 26-2-24 = 0.$$

Thus the determinant is indeed zero, confirming that the rows (or columns) of the coefficient matrix are linearly dependent. Because the system is to have infinitely many (and not zero) solutions, the same linear dependence that holds among the rows of $$A$$ must also hold among the constants $$a,\,b,\,c$$ on the right‐hand side. To discover this relation we explicitly find a non-trivial linear combination of the rows of $$A$$ that gives the zero row.

Let us look for scalars $$\alpha,\beta,\gamma$$ (not all zero) such that

$$\alpha(2,2,3)+\beta(3,-1,5)+\gamma(1,-3,2)=(0,0,0).$$

This gives the three componentwise equations

$$2\alpha+3\beta+\gamma=0,\qquad 2\alpha-\beta-3\gamma=0,\qquad 3\alpha+5\beta+2\gamma=0.$$

We solve these simultaneously. From the second equation

$$2\alpha=\beta+3\gamma\;\Longrightarrow\;\alpha=\frac{\beta+3\gamma}{2}.$$

Substituting this $$\alpha$$ into the first equation, we get

$$2\left(\frac{\beta+3\gamma}{2}\right)+3\beta+\gamma=0 \;\Longrightarrow\; (\beta+3\gamma)+3\beta+\gamma=0 \;\Longrightarrow\; 4\beta+4\gamma=0 \;\Longrightarrow\; \beta+\gamma=0 \;\Longrightarrow\; \beta=-\gamma.$$

Putting $$\beta=-\gamma$$ back into $$\alpha=\dfrac{\beta+3\gamma}{2}$$ gives

$$\alpha=\frac{-\gamma+3\gamma}{2}=\frac{2\gamma}{2}=\gamma.$$

A convenient particular choice is $$\gamma=1$$, whence $$\alpha=1$$ and $$\beta=-1$$. Therefore the linear relation among the three rows of $$A$$ is

$$(1)\times(2,2,3)+(-1)\times(3,-1,5)+(1)\times(1,-3,2)=(0,0,0).$$

Writing this compactly, we have

$$(2,2,3)\;-\;(3,-1,5)\;+\;(1,-3,2)=(0,0,0).$$

In other words,

$$\text{Row}_1-\text{Row}_2+\text{Row}_3=\mathbf{0}.$$

Exactly the same linear combination must annihilate the constant column for the augmented matrix to stay consistent. Hence we require

$$a\;-\;b\;+\;c=0.$$

Rearranging, this is equivalent to

$$-a+b-c=0\quad\text{or}\quad b-c-a=0.$$

The relation $$b-c-a=0$$ matches Option B.

Hence, the correct answer is Option B.