If \(a^{2}+b^{2}+c^{2}=2(a-b-c)-3\), then the value of \(2 a-3 b+4 c\) is (a) 1 (b) 7 (c) 2 (d) 3

Given the equation \(a^{2}+b^{2}+c^{2}=2(a-b-c)-3\), we can rewrite it as:

\[
a^2 + b^2 + c^2 – 2a + 2b + 2c + 3 = 0
\]

This equation can be rearranged to form perfect squares:

\[
(a^2 – 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0
\]

The expressions on the left-hand side can be rewritten as the squares of binomials:

\[
(a – 1)^2 + (b + 1)^2 + (c + 1)^2 = 0
\]

For the sum of squares to be zero, each square must individually be zero. Thus, we have:

\[
(a – 1)^2 = 0, \quad (b + 1)^2 = 0, \quad (c + 1)^2 = 0
\]

Solving these equations gives:

\[
a = 1, \quad b = -1, \quad c = -1
\]

Substituting these values into the expression \(2a – 3b + 4c\):

\[
2(1) – 3(-1) + 4(-1) = 2 + 3 – 4 = 1
\]

Therefore, the value of \(2a – 3b + 4c\) is \(1\).

The correct answer is (a) 1.