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Question

If a+2b+3c=4, then find the least value (to the nearest integer) of a2+b2+c2 is

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Solution

Given, a+2b+3b=4
or, (ai+bj+ck).(i+2j+3k)=4
Now taking modulus both sides we get,
a2+b2+c2.12+22+32=4
or, a2+b2+c2=414
or, a2+b2+c2=1614=87=1+17.
So least value (to the nearest integer) of (a2+b2+c2) is 1.

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