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(A) The lines $px + qy + r = 0$,$x \sin \alpha + y \cos \alpha = r$,and $x \cos \alpha - y \sin \alpha = 0$ are concurrent at point $A$.The condition

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$p^2 + q^2 = 4$

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(A) The lines $px + qy + r = 0$,$x \sin \alpha + y \cos \alpha = r$,and $x \cos \alpha - y \sin \alpha = 0$ are concurrent at point $A$.The condition for concurrency is the determinant of the coefficients being zero:$\begin{vmatrix} p & q & r \ \sin \alpha & \cos \alpha & -r \ \cos \alpha & -\sin \alpha & 0 \end{vmatrix} = 0$Expanding the determinant: $p(0 - r \sin \alpha) - q(0 + r \cos \alpha) + r(-\sin^2 \alpha - \cos^2 \alpha) = 0$$-pr \sin \alpha - qr \cos \alpha - r = 0$Since $r 
eq 0$,we divide by $-r$: $p \sin \alpha + q \cos \alpha + 1 = 0 \Rightarrow p \sin \alpha + q \cos \alpha = -1$ (Eq. $1$).The angle between $px + qy + r = 0$ and $x \sin \alpha + y \cos \alpha = r$ is $\pi / 3$.The slopes are $m_1 = -p/q$ and $m_2 = -\sin \alpha / \cos \alpha = -	an \alpha$.$	an(\pi / 3) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$ $\Rightarrow \sqrt{3} = \left| \frac{-p/q + 	an \alpha}{1 + (p/q) 	an \alpha} ight| = \left| \frac{-p \cos \alpha + q \sin \alpha}{q \cos \alpha + p \sin \alpha} ight|$.Using Eq. $1$,$q \cos \alpha + p \sin \alpha = -1$.So,$\sqrt{3} = | -p \cos \alpha + q \sin \alpha | / |-1| \Rightarrow | q \sin \alpha - p \cos \alpha | = \sqrt{3}$.Squaring both equations:$(p \sin \alpha + q \cos \alpha)^2 = (-1)^2$ $\Rightarrow p^2 \sin^2 \alpha + q^2 \cos^2 \alpha + 2pq \sin \alpha \cos \alpha = 1$$(q \sin \alpha - p \cos \alpha)^2 = (\sqrt{3})^2$ $\Rightarrow q^2 \sin^2 \alpha + p^2 \cos^2 \alpha - 2pq \sin \alpha \cos \alpha = 3$Adding these two equations:$p^2(\sin^2 \alpha + \cos^2 \alpha) + q^2(\cos^2 \alpha + \sin^2 \alpha) = 1 + 3$$p^2 + q^2 = 4$.

If $\pi / 3$ is the angle between the straight lines $px + qy + r = 0$ and $x \sin \alpha + y \cos \alpha = r$ $(r \neq 0)$ which meet at a point $A$,and the straight line $x \cos \alpha - y \sin \alpha = 0$ also passes through the point $A$,then:

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