The equations of the tangents to the hyperbola $3{\mathrm{x}}^2 – 4{\mathrm{y}}^2 = 12$ which are parallel to the line $2\mathrm{x} + \mathrm{y} + 7 = 0$ are

$$\begin{gathered} \frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{3}=1 \\ {\mathrm{a}}^2=4, {\mathrm{b}}^2=3 \\ slope of given line 2x+y+7=0 is -2 \end{gathered}$$

$\therefore$slope of required tangent $\left(\mathrm{m}\right)=-2$

Now Equation of req. tangent is $\mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2}$

$$\begin{gathered} \Rightarrow \mathrm{y}=-2\mathrm{x}\pm \sqrt{13} \\ \Rightarrow 2x+\mathrm{y}\pm \sqrt{13}=0 \end{gathered}$$