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Determinant of given quadratic = $6^{2} - 4 \times 19 = 36-76 = -40 < 0$
Thus, the roots of given quadratic are complex.

$$ Hence, the required quadratic has one root in common, it also has one complex root. But, since complex root occur in pairs, the required quadratic has both roots in common with the given quadratic. Thus, the co-efficients are in proportion.

$$

$\therefore \dfrac{a}{1} = \dfrac{b}{6} = \dfrac{c}{19} = \alpha$

$$

$\therefore a + b + c = 26 \times \alpha$

$$ For minimum integral values, of a,b,c

$$ $\alpha = 1$

$\therefore Minimum \left(a + b + c \right) = 26$

Let the common root be k.

Then the two quadratics can be written in terms of k as $\large k^2 + 6k + 19$ & $\large ak^2 + bk + c$

Subtract the two quadratics to obtain $\large (a-1)k^2 + (b-6)k + (c-19)$ .

Now clearly for k to be unique, discriminant has to be equal to zero . This gives the value of a , b & c as 6,19 & 1.

Take no knowledge of theory of equation to guess for what the author wanted,

$(x + 3)^2 + 10 = 0$

$x = - 3 \pm \sqrt{-10}$ { $\pm$ here stands for plus $Or$ minus.}

$a (- 3 \pm \sqrt{-10})^2 + b (- 3 \pm \sqrt{-10}) + c = 0$

$a (- 1 \pm -6 \sqrt{-10}) + b (- 3 \pm \sqrt{-10}) + c = 0$

We obtain two equations namely - a - 3 b + c = 0 and 6 a - b = 0.

Hence, a + b + c = a + 6 a + 19 a = 26 a = 26 with a = 1 for a minimum value wanted.

Answer: $\boxed{26}$