The given equation is $(a{x^2} + b{y^2} + c)({x^2} - 5xy + 6{y^2}) = 0$. This equation holds if either of the two factors is zero. So, the locus represented by the equation is the union of the loci represented by:

1. $ax^2 + by^2 + c = 0$
2. $x^2 - 5xy + 6y^2 = 0$

Step 1: Analyze the second equation

Let's analyze the second equation: $x^2 - 5xy + 6y^2 = 0$. This is a homogeneous equation of the second degree in $x$ and $y$. It represents a pair of straight lines passing through the origin. We can find the individual lines by factoring the expression: $x^2 - 2xy - 3xy + 6y^2 = 0$ $x(x - 2y) - 3y(x - 2y) = 0$ $(x - 2y)(x - 3y) = 0$ This represents the two straight lines $x - 2y = 0$ and $x - 3y = 0$. So, the given equation always represents at least these two straight lines.

Step 2: Analyze the first equation and evaluate the options

The first equation is $ax^2 + by^2 + c = 0$, which can be written as $ax^2 + by^2 = -c$. This represents a conic section centered at the origin. Let's evaluate each option based on the conditions provided.

A: four straight lines, when c = 0 and a, b are of the same sign If $c = 0$, the first equation becomes $ax^2 + by^2 = 0$. Since $a$ and $b$ are non-zero and have the same sign (let's say both positive), $ax^2 \ge 0$ and $by^2 \ge 0$. The sum $ax^2 + by^2$ can be zero only if $x=0$ and $y=0$. This represents a single point, the origin $(0,0)$. The total locus is the union of two straight lines and the origin. Since the two lines already pass through the origin, the locus is just the two straight lines. It is not four straight lines. Thus, option A is incorrect.

B: two straight lines and a circle, when a = b, and c is of sign opposite to that of a The second equation gives two straight lines. For the first equation, we have the conditions $a=b$ and $c$ has the opposite sign to $a$. The equation becomes $ax^2 + ay^2 + c = 0$. Since $a \neq 0$, we can divide by $a$ to get $x^2 + y^2 + \frac{c}{a} = 0$, or $x^2 + y^2 = -\frac{c}{a}$. Given that $a$ and $c$ have opposite signs, the ratio $\frac{c}{a}$ is negative. Therefore, $-\frac{c}{a}$ is positive. Let $r^2 = -\frac{c}{a} > 0$. The equation becomes $x^2 + y^2 = r^2$, which is the equation of a circle with center $(0,0)$ and radius $r = \sqrt{-c/a}$. So, the total representation is two straight lines and a circle. Thus, option B is correct.

C: two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a The second equation gives two straight lines. For the first equation, $ax^2 + by^2 = -c$. We are given that $a$ and $b$ have the same sign, and $c$ has the opposite sign. Let's assume $a > 0$ and $b > 0$. Then $c < 0$, which means $-c > 0$. The equation is $ax^2 + by^2 = K$ where $K = -c$ is a positive constant. This can be written as $\frac{x^2}{K/a} + \frac{y^2}{K/b} = 1$. Since $K>0, a>0, b>0$, the denominators are positive. This is the equation of an ellipse. A hyperbola would be formed if $a$ and $b$ had opposite signs. Thus, option C is incorrect.

D: a circle and an ellipse, when a and b are of the same sign and c is of sign opposite to that of a This option is incorrect for two reasons. First, the total locus includes two straight lines, which are not mentioned. Second, under the given conditions, the first equation $ax^2 + by^2 = -c$ represents an ellipse. A circle is a special case of an ellipse when $a=b$. The locus cannot be a circle and an ellipse simultaneously (unless we interpret it as a circle or an ellipse, which is still incomplete). Thus, option D is incorrect.