$\begin {array} {1 1} (A)\;p=-5 & \quad (B)\;p=+5 \\ (C)\;p = \pm 5 & \quad (D)\;\text{ none of the above} \end {array}$

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- If three lines may intersect at one point, then the point of intersection of lines (1) and (3) will also satisfy line (2).

The equation of the given lines are

$ \qquad 3x+y-2=0$---------(1)

$ \qquad px+2y-3=0$---------(2)

$ \qquad 2x-y-3=0$-----------(3)

Let us solve equation (1) and (3) we get,

$ \qquad 3x+y-2=0$

$ \qquad 2x-y-3=0$

$\qquad 5x-5=0 \Rightarrow x = 1$

$\qquad \therefore y = -1$

Hence the point of intersection of line (1) and line (3) is (1, -1)

Now substituting the value of $x$ and $y$ in equation (2), we get,

$ \qquad p(1)+2(-1)-3=0$

$\qquad \Rightarrow p-5=0$

$\qquad \therefore p = 5$

Hence the required value is 5.

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