If the three lines x - 3y = p,ax + 2y = q,and | vedclass

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(A) The slopes of the three lines are $m_1 = \frac{1}{3}$,$m_2 = -\frac{a}{2}$,and $m_3 = -a$.For the lines to form a right-angled triangle,the produc

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$a^2 - 9a + 18 = 0$

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(A) The slopes of the three lines are $m_1 = \frac{1}{3}$,$m_2 = -\frac{a}{2}$,and $m_3 = -a$.For the lines to form a right-angled triangle,the product of the slopes of any two perpendicular lines must be $-1$.Case $1$: $m_1 	imes m_2 = -1$ $\Rightarrow \frac{1}{3} 	imes (-\frac{a}{2}) = -1$ $\Rightarrow a = 6$.Case $2$: $m_1 	imes m_3 = -1$ $\Rightarrow \frac{1}{3} 	imes (-a) = -1$ $\Rightarrow a = 3$.Case $3$: $m_2 	imes m_3 = -1$ $\Rightarrow (-\frac{a}{2}) 	imes (-a) = -1$ $\Rightarrow \frac{a^2}{2} = -1$ (No real solution).Thus,$a = 6$ or $a = 3$ are the possible values.Testing the options for $a = 6$ and $a = 3$:For $a = 6$: $6^2 - 9(6) + 18 = 36 - 54 + 18 = 0$.For $a = 3$: $3^2 - 9(3) + 18 = 9 - 27 + 18 = 0$.Both values satisfy the equation $a^2 - 9a + 18 = 0$.

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

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