The number of integral values of $$m$$ so that the abscissa of point of intersection of lines $$3x + 4y = 9$$ and $$y = mx + 1$$ is also an integer, is:

Substituting $$y = mx + 1$$ into $$3x + 4y = 9$$ gives $$3x + 4(mx + 1) = 9$$, which simplifies to $$(3 + 4m)x = 5$$, so $$x = \frac{5}{3 + 4m}$$.

For $$x$$ to be an integer, $$(3 + 4m)$$ must be a divisor of 5. The integer divisors of 5 are $$\pm 1$$ and $$\pm 5$$.

Setting $$3 + 4m = 1$$ gives $$m = -\frac{1}{2}$$, which is not an integer. Setting $$3 + 4m = -1$$ gives $$m = -1$$, which is an integer and yields $$x = -5$$. Setting $$3 + 4m = 5$$ gives $$m = \frac{1}{2}$$, not an integer. Setting $$3 + 4m = -5$$ gives $$m = -2$$, which is an integer and yields $$x = \frac{5}{-5} = -1$$, an integer.

So the integral values of $$m$$ for which $$x$$ is also an integer are $$m = -1$$ (giving $$x = -5$$) and $$m = -2$$ (giving $$x = -1$$). There are exactly 2 such values.

The correct answer is (B) 2.