If the three lines $$x - 3y = p$$, $$ax + 2y = q$$ and $$ax + y = r$$ form a right-angled triangle then :

To solve this problem, we need to find the condition on $$a$$ such that the three given lines $$x - 3y = p$$, $$ax + 2y = q$$, and $$ax + y = r$$ form a right-angled triangle. For three lines to form a triangle, they must intersect pairwise and not be concurrent. Additionally, for the triangle to be right-angled, one pair of lines must be perpendicular, meaning the product of their slopes must be $$-1$$.

First, we find the slopes of each line by rewriting them in the slope-intercept form $$y = mx + c$$, where $$m$$ is the slope.

For the first line $$x - 3y = p$$:
Rearrange: $$x - p = 3y$$ → $$y = \frac{1}{3}x - \frac{p}{3}$$.
So, slope $$m_1 = \frac{1}{3}$$.

For the second line $$ax + 2y = q$$:
Rearrange: $$2y = -ax + q$$ → $$y = -\frac{a}{2}x + \frac{q}{2}$$.
So, slope $$m_2 = -\frac{a}{2}$$.

For the third line $$ax + y = r$$:
Rearrange: $$y = -ax + r$$.
So, slope $$m_3 = -a$$.

Since the triangle is right-angled, one pair of lines must be perpendicular. We check all possible pairs:

Case 1: Lines 1 and 2 are perpendicular.
Set $$m_1 \cdot m_2 = -1$$:
$$\frac{1}{3} \cdot \left(-\frac{a}{2}\right) = -1$$ → $$-\frac{a}{6} = -1$$.
Multiply both sides by $$-6$$: $$a = 6$$.

Case 2: Lines 1 and 3 are perpendicular.
Set $$m_1 \cdot m_3 = -1$$:
$$\frac{1}{3} \cdot (-a) = -1$$ → $$-\frac{a}{3} = -1$$.
Multiply both sides by $$-3$$: $$a = 3$$.

Case 3: Lines 2 and 3 are perpendicular.
Set $$m_2 \cdot m_3 = -1$$:
$$\left(-\frac{a}{2}\right) \cdot (-a) = -1$$ → $$\frac{a^2}{2} = -1$$.
But $$\frac{a^2}{2}$$ is always non-negative for real $$a$$, so it cannot equal $$-1$$. Thus, no solution.

So, possible values are $$a = 3$$ or $$a = 6$$. Now, we must ensure that for these values, the lines form a triangle (i.e., they are not concurrent). The lines are concurrent if the determinant of their coefficients is zero:

$$ \begin{vmatrix} 1 & -3 & -p \\ a & 2 & -q \\ a & 1 & -r \\ \end{vmatrix} = 0 $$

Expanding the determinant:
$$1 \cdot (2 \cdot (-r) - (-q) \cdot 1) - (-3) \cdot (a \cdot (-r) - (-q) \cdot a) + (-p) \cdot (a \cdot 1 - 2 \cdot a)$$
$$= 1 \cdot (-2r + q) + 3 \cdot (-a r + a q) - p \cdot (-a)$$
$$= -2r + q - 3a r + 3a q + a p$$
Set to zero: $$a p + 3a q - 3a r + q - 2r = 0$$.

Since $$p$$, $$q$$, and $$r$$ are arbitrary constants, we can choose values such that this expression is not zero. For example:

• For $$a = 3$$, choose $$p = 1$$, $$q = 1$$, $$r = 1$$: Lines are $$x - 3y = 1$$, $$3x + 2y = 1$$, $$3x + y = 1$$. Solving the last two: $$3x + 2y = 1$$ and $$3x + y = 1$$ gives $$y = 0$$, $$x = \frac{1}{3}$$. Plugging into first: $$\frac{1}{3} - 0 = \frac{1}{3} \neq 1$$, so not concurrent.
• For $$a = 6$$, choose $$p = 1$$, $$q = 1$$, $$r = 1$$: Lines are $$x - 3y = 1$$, $$6x + 2y = 1$$, $$6x + y = 1$$. Solving the last two: $$6x + 2y = 1$$ and $$6x + y = 1$$ gives $$y = 0$$, $$x = \frac{1}{6}$$. Plugging into first: $$\frac{1}{6} - 0 = \frac{1}{6} \neq 1$$, so not concurrent.

Thus, for both $$a = 3$$ and $$a = 6$$, we can choose $$p$$, $$q$$, and $$r$$ such that the lines form a triangle. Since one pair is perpendicular in each case, the triangle is right-angled.

The values $$a = 3$$ and $$a = 6$$ must satisfy a quadratic equation. Checking the options:

• Option A: $$a^2 - 9a + 18 = 0$$ → $$(a - 3)(a - 6) = 0$$, roots $$a = 3, 6$$.
• Option B: $$a^2 - 6a - 12 = 0$$ → discriminant $$36 + 48 = 84$$, roots $$\frac{6 \pm \sqrt{84}}{2} = 3 \pm \sqrt{21}$$, not 3 or 6.
• Option C: $$a^2 - 6a - 18 = 0$$ → discriminant $$36 + 72 = 108$$, roots $$\frac{6 \pm \sqrt{108}}{2} = 3 \pm 3\sqrt{3}$$, not 3 or 6.
• Option D: $$a^2 - 9a + 12 = 0$$ → discriminant $$81 - 48 = 33$$, roots $$\frac{9 \pm \sqrt{33}}{2}$$, not 3 or 6.

Hence, the correct answer is Option A.