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The least value of the expression x^2+4y...

The least value of the expression `x^2+4y^2+3z^2-2x-12 y-6z+14` is `3` b. no least value`` c. `0` d. none of these

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Let `f(x, y, z) = x^(2) + 4y^(2) + 3z^(2) - 2x - 12y - 6z + 14`
=` (x - 1)^(2) + (2y - 3)^(2) + 3(z - 1 )^(2) + 1`
For least value of f(x, y, z),
`x - 1 = 0, 2y- 3 = 0 and z - 1 = 0`
`therefore x = 1, y = 3//2, z = 1`
Hence, the least value of f(x, y, z) is `f(1, 3//2,1) = 1.`
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