If $$a$$, $$b$$, $$c$$ are non-zero real numbers and if the system of equations
$$(a-1)x = y + z$$
$$(b-1)y = x + z$$
$$(c-1)z = x + y$$
has a non-trivial solution, then $$ab + bc + ca$$ equals:

The given equations are

$$\begin{aligned} (a-1)x &= y+z,\\ (b-1)y &= x+z,\\ (c-1)z &= x+y. \end{aligned}$$

Bring every term to the left so that each equation is of the form $$A_{11}x+A_{12}y+A_{13}z=0.$$

$$\begin{aligned} (a-1)x-y-z &= 0,\\ -x+(b-1)y-z &= 0,\\ -x-y+(c-1)z &= 0. \end{aligned}$$

Hence the coefficient matrix of the system is

$$ \begin{vmatrix} a-1 & -1 & -1\\ -1 & b-1 & -1\\ -1 & -1 & c-1 \end{vmatrix}. $$

For a non-trivial solution the determinant of this matrix must be zero.

Introduce the short-hand $$p=a-1,\;q=b-1,\;r=c-1.$$ The determinant becomes

$$ \Delta= \begin{vmatrix} p & -1 & -1\\ -1 & q & -1\\ -1 & -1 & r \end{vmatrix}. $$

Expand along the first row (sign pattern $$+\;-\;+\,$$):

$$ \begin{aligned} \Delta &= p\Big(qr-(-1)(-1)\Big) \;-\;(-1)\Big((-1)r-(-1)(-1)\Big) \;+\;(-1)\Big((-1)(-1)-q(-1)\Big)\\ &= p(qr-1) +\big(-r-1\big) -\big(1+q\big). \end{aligned} $$

Simplify the last line:

$$ \Delta = pqr - p - q - r - 2 = pqr - p - (q+r) - 2. $$

Rewrite $$- (q+r) - 2$$ in terms of $$b,c$$:

$$ -(q+r)-2 = -(b-1) - (c-1) -2 = -b-c. $$

Therefore

$$ \Delta = pqr - p - b - c. $$

Substitute back $$p=a-1$$ and expand:

$$ \begin{aligned} \Delta &= (a-1)(b-1)(c-1) - (a-1) - (b+c)\\ &= \big(abc -ab-bc-ca +a+b+c -1\big) -a +1 -b -c\\ &= abc -ab -bc -ca. \end{aligned} $$

For a non-trivial solution we need $$\Delta = 0$$, hence

$$ abc - (ab+bc+ca)=0 \quad\Longrightarrow\quad ab+bc+ca = abc. $$

This matches Option C.

Final answer: $$ab+bc+ca = abc\;.$$