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(D) The given equation of the line is $3 \cos 	heta \cdot x + 4 \sin 	heta \cdot y = 12$.Dividing by $12$,we get the intercept form:$\frac{x}{4 / \c

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$12$

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(D) The given equation of the line is $3 \cos 	heta \cdot x + 4 \sin 	heta \cdot y = 12$.Dividing by $12$,we get the intercept form:$\frac{x}{4 / \cos 	heta} + \frac{y}{3 / \sin 	heta} = 1$.This line intersects the coordinate axes at $A\left(\frac{4}{\cos 	heta}, 0ight)$ and $B\left(0, \frac{3}{\sin 	heta}ight)$.The area of $	riangle OAB$ is given by:$\Delta = \frac{1}{2} 	imes |OA| 	imes |OB| = \frac{1}{2} 	imes \left| \frac{4}{\cos 	heta} ight| 	imes \left| \frac{3}{\sin 	heta} ight| = \frac{6}{|\sin 	heta \cos 	heta|} = \frac{12}{|\sin 2 	heta|}$.For the area to be minimum,$|\sin 2 	heta|$ must be maximum. Since the maximum value of $|\sin 2 	heta|$ is $1$,the minimum area is:$\Delta_{\min} = \frac{12}{1} = 12$.

The minimum area of the triangle formed by the variable line $3 \cos \theta \cdot x + 4 \sin \theta \cdot y = 12$ and the coordinate axes is

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