If the equation $kx^2 - 5xy + 6y^2 + x - 3y = 0$ represents a pair of straight lines, then the co-ordinates of their point of intersection are

Comparing the given equation with the general equation of second degree $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$, we get:

$a = k$, $h = -\dfrac{5}{2}$, $b = 6$, $g = \dfrac{1}{2}$, $f = -\dfrac{3}{2}$, $c = 0$

For the equation to represent a pair of straight lines, the condition is $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.

Substituting the values:

$0 + 2\left(-\dfrac{3}{2}\right)\left(\dfrac{1}{2}\right)\left(-\dfrac{5}{2}\right) - k\left(-\dfrac{3}{2}\right)^2 - 6\left(\dfrac{1}{2}\right)^2 - 0 = 0$

$\dfrac{15}{4} - \dfrac{9k}{4} - \dfrac{6}{4} = 0$

$\dfrac{9}{4} - \dfrac{9k}{4} = 0 \Rightarrow k = 1$

The equation of the pair of straight lines is $x^2 - 5xy + 6y^2 + x - 3y = 0$.

Factorizing the equation:

$(x - 3y)(x - 2y) + 1(x - 3y) = 0$

$(x - 3y)(x - 2y + 1) = 0$

The equations of the individual lines are $x - 3y = 0$ and $x - 2y + 1 = 0$.

Solving these two equations, we substitute $x = 3y$ into the second equation:

$3y - 2y + 1 = 0 \Rightarrow y = -1$

Substituting $y = -1$ back into $x = 3y$, we get $x = -3$.

The coordinates of the point of intersection are $(-3, -1)$.

Answer: $(-3, -1)$