If the equation $\mathrm{12 x^2+7 x y-p y^2-18 x+g y+6=0}$ represents two perpendicular lines, then the values of $\mathrm{\mathrm{p} \: and \: \mathrm{q}}$ are  Option: 1 $12,1$  Option: 2 $12,-1$Option: 3 $-12,1$Option: 4 $-12,-1$

$\mathrm{12-\mathrm{p}=0 \quad \mathrm{p}=12}$ (since, lines are perpendicular)

$\mathrm{\text { and }\left|\begin{array}{ccc}12 & \frac{7}{2} & -9 \\ \frac{7}{2} & -12 & \frac{q}{2} \\ \end{array}\right|=0}$

$\mathrm{ 12\left(-72-\frac{q^2}{4}\right)-\frac{7}{2}\left(21+\frac{9 q}{2}\right)-9\left(\frac{7 q}{4}-108\right)=0 }$

$\mathrm{ \Rightarrow \quad-864-3 q^2-\frac{147}{2}-\frac{63 q}{4}-\frac{63 q}{4}+972=0 }$

$\mathrm{ \Rightarrow \quad 108-3 q^2-\frac{147}{2}-\frac{63 q}{2}=0 }$

$\mathrm{ \Rightarrow \quad 6 q^2+63 q-69=0 }$

$\mathrm{ \Rightarrow \quad 2 q^2+21 q-23=0 }$

$\mathrm{\Rightarrow \quad (q-1)\left(2 q+\frac{23}{2}\right)=0 }$

$\mathrm{\Rightarrow \quad q=1 \text { or } q=-23 / 2 }$

$\mathrm{\therefore p=12 \text { and } q=1 }$

Hence option 1 is correct.