If a+x=b+y=c+z+1, where a, b, c, x, y, z are | vedclass

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(B) Given $a+x=b+y=c+z+1=k$ (let).Let $\Delta = \left|\begin{array}{lll}x & a+y & x+a \ y & b+y & y+b \ z & c+y & z+c\end{array}ight|$.Applying $C

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$y(a-b)$

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(B) Given $a+x=b+y=c+z+1=k$ (let).Let $\Delta = \left|\begin{array}{lll}x & a+y & x+a \ y & b+y & y+b \ z & c+y & z+c\end{array}ight|$.Applying $C_3 ightarrow C_3 - C_1$:$\Delta = \left|\begin{array}{lll}x & a+y & a \ y & b+y & b \ z & c+y & c\end{array}ight|$.Applying $C_2 ightarrow C_2 - C_3$:$\Delta = \left|\begin{array}{lll}x & y & a \ y & y & b \ z & y & c\end{array}ight|$.Taking $y$ common from $C_2$:$\Delta = y \left|\begin{array}{lll}x & 1 & a \ y & 1 & b \ z & 1 & c\end{array}ight|$.From the given equations: $x = k-a, y = k-b, z = k-c-1$.Substituting these values:$\Delta = y \left|\begin{array}{lll}k-a & 1 & a \ k-b & 1 & b \ k-c-1 & 1 & c\end{array}ight|$.Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$\Delta = y \left|\begin{array}{lll}k-a & 1 & a \ b-a & 0 & b-a \ c-a-1 & 0 & c-a\end{array}ight|$.Expanding along $C_2$:$\Delta = y(-1) [(b-a)(c-a) - (c-a-1)(b-a)]$$\Delta = -y(b-a) [c-a - (c-a-1)]$$\Delta = -y(b-a) [1] = y(a-b)$.

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

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