If $a^{2} + b^{2} + c^{2} = 2 (a - b - c) - 3$ then find the value of $2 a - 3 b + 4 c$ .

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Solution

Given, $a^{2} + b^{2} + c^{2} = 2 (a - b - c) - 3$
or, $a^{2} - 2 a + 1 + b^{2} + 2 b + 1 + c^{2} + 2 c + 1 = 0$
or, $(a - 1)^{2} + (b + 1)^{2} + (c + 1)^{2} = 0$
From this equation we get, $a - 1 = 0, b + 1 = 0, c + 1 = 0$ or, $a = 1, b = - 1, c = - 1$ .
Then
$2 a - 3 b + 4 c$
$= 2.1 - 3. (- 1) + 4. (- 1)$
$= 5 - 4$
$= 1$ .

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If $a^{2} + b^{2} = 14$ and $a b = 3$ , find the value of $2 (a + b)^{2} - 5 (a - b)^{2}$ .

\frac{a}{2} = \frac{b}{3} = \frac{c}{4} = \frac{2 a - 3 b + 4 c}{p}

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