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# If $D=\begin{vmatrix}1&1&1\\1&1+x&1\\1&1&1+y\end{vmatrix}$ for $x\neq 0,y\neq 0$ then $D$ is

$\begin{array}{1 1}(a)\;divisible\;by\;x\;but\;not\;y\\(b)\;divisible\;by\;y\;but\;not\;x\\(c)\;divisible\;by\;neither\;x\;nor\;y\\(d)\;divisible\;by\;both\;x\;and\;y\end{array}$

Given $D=\begin{bmatrix}1&1&1\\1&1+x&1\\1&1&1+y\end{bmatrix}$
Apply $R_2\rightarrow R_2-R_1$ and $R\rightarrow R_3-R_1$
$\therefore D=\begin{vmatrix}1 &1&1\\0&x&0\\0&0&y\end{vmatrix}=xy$
Hence D is divisible by both x and y
Hence (d) is the correct answer.