Correct Option: B

Given the equations:1. $a + b + c = 6$2. $a^2 + b^2 + c^2 = 14$From equation (1), we can express $a+b$ in terms of c: $a + b = 6 - c$.From equation (2), we can express $a^2+b^2$ in terms of c: $a^2 + b^2 = 14 - c^2$.We know the algebraic identity: $(a+b)^2 = a^2 + b^2 + 2ab$.Substitute the expressions for $a+b$ and $a^2+b^2$ into this identity:$(6 - c)^2 = (14 - c^2) + 2ab$Expand the left side:$36 - 12c + c^2 = 14 - c^2 + 2ab$Now, isolate $2ab$:$2ab = 36 - 12c + c^2 - 14 + c^2$$2ab = 2c^2 - 12c + 22$Divide by 2 to find $ab$:$ab = c^2 - 6c + 11$For real numbers a and b to exist, they must be the roots of a quadratic equation. This quadratic equation can be formed as $t^2 - (a+b)t + ab = 0$.Substitute the expressions for $a+b$ and $ab$:$t^2 - (6 - c)t + (c^2 - 6c + 11) = 0$For this quadratic equation to have real roots (a and b), its discriminant (D) must be greater than or equal to zero ($D \ge 0$).The discriminant is $D = B^2 - 4AC$, where $A=1$, $B=-(6-c)$, and $C=(c^2 - 6c + 11)$.$D = (-(6-c))^2 - 4(1)(c^2 - 6c + 11) \ge 0$$(6-c)^2 - 4(c^2 - 6c + 11) \ge 0$Expand and simplify the inequality:$(36 - 12c + c^2) - (4c^2 - 24c + 44) \ge 0$$36 - 12c + c^2 - 4c^2 + 24c - 44 \ge 0$$-3c^2 + 12c - 8 \ge 0$Multiply the entire inequality by -1 and reverse the inequality sign:$3c^2 - 12c + 8 \le 0$To find the values of c that satisfy this quadratic inequality, we first find the roots of the corresponding quadratic equation $3c^2 - 12c + 8 = 0$ using the quadratic formula $c = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$:$c = \frac{12 \pm \sqrt{(-12)^2 - 4(3)(8)}}{2(3)}$$c = \frac{12 \pm \sqrt{144 - 96}}{6}$$c = \frac{12 \pm \sqrt{48}}{6}$Simplify $\sqrt{48}$ as $\sqrt{16 \times 3} = 4\sqrt{3}$:$c = \frac{12 \pm 4\sqrt{3}}{6}$Divide the numerator and denominator by 2:$c = 2 \pm \frac{2\sqrt{3}}{3}$Let the roots be $c_1 = 2 - \frac{2\sqrt{3}}{3}$ and $c_2 = 2 + \frac{2\sqrt{3}}{3}$.Since the coefficient of $c^2$ in $3c^2 - 12c + 8$ is positive (3 > 0), the parabola opens upwards. For the inequality $3c^2 - 12c + 8 \le 0$ to hold, c must lie between or be equal to its roots.Thus, the range of c is $\left[2 - \frac{2\sqrt{3}}{3}, 2 + \frac{2\sqrt{3}}{3}\right]$.This corresponds to Option B.