Question : If a, b, c are real and $a^{2}+b^{2}+c^{2}=2(a-b-c)-3, $ then the value of $2a-3b+4c$ is:
Option 1: –1
Option 2: 0
Option 3: 1
Option 4: 2
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Correct Answer: 1
Solution :
Given: $a^{2}+b^{2}+c^{2}=2(a-b-c)-3$.
We can simplify the above equation as follows:
⇒ $a^{2}+b^{2}+c^{2}=2a-2b-2c-3$
⇒ $a^{2}-2a+1+b^{2}+2b+1+c^{2}+2c+1=0$
⇒ $(a-1)^{2}+(b+1)^{2}+(c+1)^{2}=0$
$\therefore a=1, b=-1,c=-1$
Putting the values of $a, b, c$ in $2a-3b+4c$, we get:
$2+3-4=1$
Hence, the correct answer is 1.
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