Given pair of lines is $6x^2+xy-y^2=0$.

$\Rightarrow 6x^2+3xy-2xy-y^2=0$

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

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

$\therefore$The lines are $3x-y=0$ and $2x+y=0$.

If $3x-y=0$ is a common line to $3x^2-axy-y^2=0$.

If $3x-y=0\Rightarrow 3{\mathrm{x}}^2-axy-y^2=0$

If $3x-y=0\Rightarrow y=3x$

Substitute $y=3x$

$\Rightarrow 3x^2-ax\left(3x\right)-{\left(3x\right)}^2=0$

$\Rightarrow 3x^2-3a{x}^2-9x^2=0$

$\Rightarrow -3a{x}^2=6x^2$

$\Rightarrow a=-2$

Substitute $2x+y=0\Rightarrow y=-2x$

$\Rightarrow 3x^2-ax\left(-2x\right)-{\left(-2x\right)}^2=0$

$\Rightarrow 3x^2+2a{x}^2-4x^2=0$

$\Rightarrow 2a=1$

$\Rightarrow a=\frac{1}{2}$