Slopes of line respectively by
$3 x^{2} + k x y - y^{2} = 0$ differ by $4$

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Solution

According to the problem
Let us consider the equation
$3 x^{2} + k x y - y^{2} = 0$
$m_{1} m_{2} = \frac{k}{- 1} = - k$

Therefore,
$m_{1} m_{2} = - 3$

Hence,
$(m_{1} - m_{2})^{2} = (m_{1} + m_{2})^{2} - 4 m_{1} m_{2}$

Implies that
$k^{2} + 12 = 16$

Since, $m_{1} - m_{2} = 4$
$k = \pm 2$

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