Find the least value of the expression $x^{2} + 4 y^{2} + 3 z^{2} - 2 x - 12 y - 6 z + 14$ .

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Solution

Que. least value
$x^{2} + 4 y^{2} + 32^{2} - 2 x - 12 y - 62 + 14$
$\to= (x^{2} - 2 x + 1) + 4 (y^{2} - 2 \times \frac{3}{2} \times 4 + \frac{9}{4}) + 3 (2^{2} - 22 + 1) + 14 - 1 - 9 - 3$
$= (x + 1)^{2} + 4 (y - 3 / 2)^{2} + 3 (2 - 1)^{2} + (2 - 1)^{2} + 1$
since the first three term of this can not be
negative. $(∵ (- a)^{2} = a^{2})$
the smallest value of Question is 1
so the answer is 1

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