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यदि $\left|\begin{array}{ccc} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array}\right|=( a + b + c )$ $( x + a + b + c )^{2}, x \neq 0$ तथा $a + b + c \neq 0$ हो, तो $x$ बराबर है
$abc$
$ - 2 \left( {a + b + c} \right)$
$ 2 \left( {a + b + c} \right)$
$ - \left( {a + b + c} \right)$
Solution
${R_1} \to {R_1} + {R_2} + {R_3}$
$\left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}
1&1&1\\
{2b}&{b – a – c}&{2b}\\
{2c}&{2c}&{c – a – b}
\end{array}} \right|$
${c_3} \to {c_3} – {c_1},{c_2} \to {c_2} – {c_1}$
$ = \left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}
1&0&0\\
{2b}&{ – \left( {a + b + c} \right)}&0\\
{2c}&0&{ – \left( {a + b + c} \right)}
\end{array}} \right|$
$ = {\left( {a + b + c} \right)^3}$
$ = \left( {a + b + c} \right){\left( {x + a + b + c} \right)^2}$
