Equation of pair of straight lines drawn through (1, 1) and perpendicular to the pair of lines $3x^2-7xy-2y^2=0$ is

Given: 

Equation of pair of straight lines drawn through (1,1) and perpendicular to the pair of lines

$3x^2-7xy-2y^2=0$is given by

Formula: Equation of pair of perpendicular to the pair of lines $a{x}^2+2hxy+b{y}^2=0$ and passing through origin is given by $b{x}^2-2hxy+a{y}^2=0$

Note: We don't know whether the pair of perpendicular passing through origin or not but we know it is passing through (1,1) so, equation must be in the form of 

$b{x}^2-2hxy+a{y}^2+2gx+2fy+c=0$

Formula: Equation of pair of perpendicular to the pair of lines $a{x}^2+2hxy+b{y}^2=0$ and passing through the point $(x_1,y_1)$ is given by

$$\begin{gathered} b(x-x{)}^2-2h(x-x)\left(y-y_1\right)+a{(y-y_1)}^2=0 a=3,b=-2,2h=-7 \\ -2(x-1{)}^2+7(x-1)(y-1)+3(y-1{)}^2=0 \\ -2\left(x^2-2x+1\right)+7(xy-x-y+1)+3\left(y^2-2y+1\right)=0 \\ -2x^2+4x-2+7xy-7x-7y+7+3y^2-6y+3=0 \\ -2x^2+7xy+3y^2-3x-13y+8=0 \end{gathered}$$

Correct option is 4.