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Let P be any point on the line $x - y + 3 = 0$ and A be a fixed point $(3, 4) .$ If the family of lines given by the equations $(3 \sec θ + 5 \cos e c θ) x + (7 \sec θ - 3 \cos e c θ) y + 11 (\sec θ - \cos e c θ) = 0$ are concurrent at a point B for all permissible value of $θ$ , then:

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Sum of the abscissa and ordinate of point B is equal to $- 1$

Product of the abscissa and ordinate of point B is equal to $- 2$

Maximum value of $| P A - P B |$ is $2 \sqrt{10}$

Minimum value of $P A + P B$ is $2 \sqrt{34}$

answer is A, B, C.

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Detailed Solution

$(3 \sec θ + 5 \cos e c θ) x + (7 \sec θ - 3 \cos e c θ) y + 11 (\sec θ - \cos e c θ) = 0 \sec θ (3 x + 7 y + 11) + \cos e c θ (5 x - 3 y - 11) = 0$

Hence, family of lines are concurrent at the point of intersection of $3 x + 7 y + 11 = 0$ and $5 x - 3 y - 11 = 0$
Hence, point B is $(1, - 2)$
Now proceed.

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