If `a+2b+3c=4,`
then find the least value
of `a^2+b^2+c^2dot`
A
`-2`
B
`2`
C
`8`
D
`-14`
Text Solution
Verified by Experts
The correct Answer is:
D
`(d)` `ab+bc+ca=((a+b+c)^(2)-(a^(2)+b^(2)+c^(2)))/(2)=183` Hence `a`, `b`, `c` are the roots of the equation. `t^(3)-24t^(2)+183t-440=0` `implies(t-5)(t-8)(t-11)=0` Thus `{a,b,c}={5,8,11}`