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(C) The equation of a line passing through $(2, 3)$ with slope $m$ is $y - 3 = m(x - 2)$,which simplifies to $y = mx + (3 - 2m)$.The $x$-intercept $(A

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$3$

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(C) The equation of a line passing through $(2, 3)$ with slope $m$ is $y - 3 = m(x - 2)$,which simplifies to $y = mx + (3 - 2m)$.The $x$-intercept $(A)$ is found by setting $y = 0$: $0 = mx + 3 - 2m \Rightarrow x = \frac{2m - 3}{m}$.The $y$-intercept $(B)$ is found by setting $x = 0$: $y = 3 - 2m$.The area of the triangle formed with the coordinate axes is $\frac{1}{2} |x_{intercept}| |y_{intercept}| = 12$.$\frac{1}{2} |\frac{2m - 3}{m}| |3 - 2m| = 12$.Since $(3 - 2m)^2 = (2m - 3)^2$,we have $\frac{(2m - 3)^2}{|m|} = 24$.Case $1$: $m > 0$,then $(2m - 3)^2 = 24m$ $\Rightarrow 4m^2 - 12m + 9 = 24m$ $\Rightarrow 4m^2 - 36m + 9 = 0$. The discriminant $D = (-36)^2 - 4(4)(9) = 1296 - 144 = 1152 > 0$,so there are $2$ distinct positive values for $m$.Case $2$: $m Total number of possible lines = $2 + 1 = 3$.

Column $I$	Column $II$
$(a)$ Proliferative Phase	$(i)$ Breakdown of endometrium
$(b)$ Secretory Phase	$(ii)$ Follicular phase
$(c)$ Menstruation	$(iii)$ Luteal Phase

The number of possible straight lines passing through $(2, 3)$ and forming a triangle with the coordinate axes,whose area is $12 \, sq. \, units$,is

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Match the items given in Column $I$ with those in Column $II$ and select the correct option given below:

Column $I$ Column $II$

$(a)$ Proliferative Phase $(i)$ Breakdown of endometrium

$(b)$ Secretory Phase $(ii)$ Follicular phase

$(c)$ Menstruation $(iii)$ Luteal Phase

If $a \neq p, b \neq q, c \neq r$ and $\left|\begin{array}{ccc}p & b & c \\ p+a & q+b & 2c \\ a & b & r\end{array}\right|=0$,then $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is equal to :

$(\sqrt{3}+i)^7+(\sqrt{3}-i)^7$ is equal to (in $sqrt{3}$)

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