If {a^2} + {b^2} + {c^2} + ab + bc + ca leq 0 | vedclass

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(D) Given the inequality: ${a^2} + {b^2} + {c^2} + ab + bc + ca \leq 0$.
Multiplying by $2$,we get: $2a^2 + 2b^2 + 2c^2 + 2ab + 2bc + 2ca \leq 0$.
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(D) Given the inequality: ${a^2} + {b^2} + {c^2} + ab + bc + ca \leq 0$.
Multiplying by $2$,we get: $2a^2 + 2b^2 + 2c^2 + 2ab + 2bc + 2ca \leq 0$.
This can be rewritten as: $(a^2 + 2ab + b^2) + (b^2 + 2bc + c^2) + (c^2 + 2ca + a^2) \leq 0$.
$(a+b)^2 + (b+c)^2 + (c+a)^2 \leq 0$.
Since the sum of squares of real numbers is non-negative,the only way the sum is $\leq 0$ is if each term is zero:
$a+b=0, b+c=0, c+a=0$.
Solving these equations,we get $a=0, b=0, c=0$.
Now,substitute $a=0, b=0, c=0$ into the determinant:
$\Delta = \left| {\begin{array}{*{20}{c}} {{(0+0+0)}^2} & {{0^2} + {0^2}} & 1 \ 1 & {{(0+0+2)}^2} & {{0^2} + {0^2}} \ {{0^2} + {0^2}} & 1 & {{(0+0+2)}^2} \end{array}} ight| = \left| {\begin{array}{*{20}{c}} 0 & 0 & 1 \ 1 & 4 & 0 \ 0 & 1 & 4 \end{array}} ight|$.
Expanding along the first row:
$\Delta = 0(16-0) - 0(4-0) + 1(1-0) = 1$.

If ${a^2} + {b^2} + {c^2} + ab + bc + ca \leq 0$ for all $a, b, c \in R$,then find the value of the determinant $\left| {\begin{array}{*{20}{c}} {{(a + b + c)}^2} & {{a^2} + {b^2}} & 1 \\ 1 & {{(b + c + 2)}^2} & {{b^2} + {c^2}} \\ {{c^2} + {a^2}} & 1 & {{(c + a + 2)}^2} \end{array}} \right|$.

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