Let the set of all values of k in R such that | vedclass

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(A) Let $z = x+iy$. The equation is $z\bar{z} + z(2+i) + k(2+3i) = 0$.
Substituting $z = x+iy$ and $\bar{z} = x-iy$, we get $(x^2 + y^2) + (x+iy)(2+i

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(A) Let $z = x+iy$. The equation is $z\bar{z} + z(2+i) + k(2+3i) = 0$.
Substituting $z = x+iy$ and $\bar{z} = x-iy$, we get $(x^2 + y^2) + (x+iy)(2+i) + 2k + 3ki = 0$.
Expanding the terms: $x^2 + y^2 + 2x + ix + 2iy - y + 2k + 3ki = 0$.
Separating real and imaginary parts:
Real part: $x^2 + y^2 + 2x - y + 2k = 0$
Imaginary part: $x + 2y + 3k = 0 \Rightarrow x = -2y - 3k$.
Substitute $x$ into the real part equation:
$(-2y-3k)^2 + y^2 + 2(-2y-3k) - y + 2k = 0$
$4y^2 + 12yk + 9k^2 + y^2 - 4y - 6k - y + 2k = 0$
$5y^2 + y(12k - 5) + 9k^2 - 4k = 0$.
For $y$ to be real, the discriminant $D \ge 0$:
$D = (12k - 5)^2 - 4(5)(9k^2 - 4k) \ge 0$
$144k^2 - 120k + 25 - 180k^2 + 80k \ge 0$
$-36k^2 - 40k + 25 \ge 0 \Rightarrow 36k^2 + 40k - 25 \le 0$.
The roots of $36k^2 + 40k - 25 = 0$ are $k = \frac{-40 \pm \sqrt{1600 - 4(36)(-25)}}{72} = \frac{-40 \pm \sqrt{5200}}{72}$.
Thus, $\alpha + \beta = -\frac{40}{36} = -\frac{10}{9}$.
Therefore, $9(\alpha + \beta) = 9(-\frac{10}{9}) = -10$.

List-$I$	List-$II$
$(i)$ $a \bar{a}$	$(A)$ $-\frac{\pi}{3}$
$(ii)$ $\arg \left(\frac{1}{\bar{a}}\right)$	$(B)$ $-i \sqrt{3}$
$(iii)$ $a - \bar{a}$	$(C)$ $2i / \sqrt{3}$
$(iv)$ $\operatorname{Im}\left(\frac{4}{3a}\right)$	$(D)$ $1$
	$(E)$ $\pi / 3$
	$(F)$ $\frac{2}{\sqrt{3}}$

Let the set of all values of $k \in R$ such that the equation $z(\bar{z} + 2 + i) + k(2 + 3i) = 0, z \in C$, has at least one solution, be the interval $[\alpha, \beta]$. Then $9(\alpha + \beta)$ is equal to:

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