If $$a + x = b + y = c + z + 1$$, where $$a, b, c, x, y, z$$ are non-zero distinct real numbers, then $$\begin{vmatrix} x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c \end{vmatrix}$$ is equal to:

We are told that the three sums are equal :

$$a+x=b+y=c+z+1.$$

Put the common value equal to $$k$$. Then

$$a+x=k,\qquad b+y=k,\qquad c+z+1=k.$$

From these we can express $$x,y,z$$ entirely through the independent symbols $$a,b,c$$ and the constant $$k$$ :

$$x=k-a,\qquad y=k-b,\qquad z=k-1-c.$$

Now substitute these expressions in the given determinant

$$ \Delta=\begin{vmatrix} x & a+y & x+a\\[4pt] y & b+y & y+b\\[4pt] z & c+y & z+c \end{vmatrix}. $$

After substitution each entry becomes

$$ \begin{aligned} x &=k-a,\\ a+y &=a+(k-b)=k+a-b,\\ x+a &=(k-a)+a=k,\\[2pt] y &=k-b,\\ b+y &=b+(k-b)=k,\\ y+b &=(k-b)+b=k,\\[2pt] z &=k-1-c,\\ c+y &=c+(k-b)=k+c-b,\\ z+c &=(k-1-c)+c=k-1. \end{aligned} $$

Therefore the determinant acquires the concrete numerical shape

$$ \Delta=\begin{vmatrix} k-a & k+a-b & k \\[4pt] k-b & k & k \\[4pt] k-1-c & k+c-b & k-1 \end{vmatrix}. $$

Perform the row operation $$R_1\to R_1-R_2$$ (row-1 minus row-2). Row operations of the form $$R_i\to R_i+\lambda R_j$$ do not alter the value of a determinant. We get

$$ \Delta=\begin{vmatrix} (b-a) & (a-b) & 0 \\[4pt] k-b & k & k \\[4pt] k-1-c & k+c-b & k-1 \end{vmatrix}. $$

Next isolate the common factor $$b-a$$ from the first row:

$$ \Delta=(b-a)\begin{vmatrix} 1 & -1 & 0 \\[4pt] k-b & k & k \\[4pt] k-1-c & k+c-b & k-1 \end{vmatrix}. $$

Call the remaining determinant $$D_1$$, so $$\Delta=(b-a)D_1$$ where

$$ D_1=\begin{vmatrix} 1 & -1 & 0 \\[4pt] k-b & k & k \\[4pt] k-1-c & k+c-b & k-1 \end{vmatrix}. $$

The third element of the first row is already zero, so expand $$D_1$$ along that row (Laplace expansion for a $$3\times3$$ determinant):

$$ D_1 =1\begin{vmatrix} k & k \\[4pt] k+c-b & k-1 \end{vmatrix} -(-1)\begin{vmatrix} k-b & k \\[4pt] k-1-c & k-1 \end{vmatrix}. $$

Recall the $$2\times2$$ determinant formula $$\begin{vmatrix}p&q\\r&s\end{vmatrix}=ps-qr.$$ Evaluate both minors separately.

First minor :

$$ \begin{vmatrix} k & k \\[4pt] k+c-b & k-1 \end{vmatrix} =k(k-1)-k(k+c-b)=-k\big[(k+c-b)-(k-1)\big]=-k(c-b+1). $$

Second minor :

$$ \begin{vmatrix} k-b & k \\[4pt] k-1-c & k-1 \end{vmatrix} =(k-b)(k-1)-k(k-1-c) =-(k-b)-k(b-c-1)=-k+b-k(b-c-1). $$

Simplify that second expression carefully:

$$ -k+b-k(b-c-1) =-k+b-kb+kc+k =-kb+kc+b. $$

Now assemble $$D_1$$ :

$$ \begin{aligned} D_1 &= 1\big[-k(c-b+1)\big] +1\big[-kb+kc+b\big]\\[4pt] &=-k(c-b+1)-kb+kc+b. \end{aligned} $$

Observe that the two terms involving $$kc$$ cancel: $$-k(c-b+1)=-kc+kb-k$$, so

$$ D_1=(-kc+kb-k)-kb+kc+b=-k+b. $$

Therefore

$$D_1=-k+b.$$

But $$y=k-b$$, hence $$-k+b=-(k-b)=-y.$$ So

$$D_1=-y.$$

Returning to $$\Delta=(b-a)D_1$$ gives

$$ \Delta=(b-a)(-y)=y(a-b). $$

Thus the value of the determinant is $$y(a-b)$$, which coincides with Option B.

Hence, the correct answer is Option B.