The angle between the lines $$2x = 3y = -z$$ and $$6x = -y = -4z$$ is

A line written in symmetric (proportional) form $$\dfrac{x - x_0}{a} = \dfrac{y - y_0}{b} = \dfrac{z - z_0}{c}$$ has direction ratios (d.r.’s) $$a,\,b,\,c$$.
Therefore, to find the angle between two given lines, first extract their direction ratios, then apply the dot-product formula.

Line 1: $$2x = 3y = -z$$
Put the common value as $$\lambda$$: $$2x = 3y = -z = \lambda$$.
Hence $$x = \dfrac{\lambda}{2},\; y = \dfrac{\lambda}{3},\; z = -\lambda$$.
So the direction ratios are proportional to $$\left(\dfrac{1}{2},\,\dfrac{1}{3},\,-1\right)$$.
Multiply by $$6$$ to clear denominators: first line d.r.’s $$\mathbf{a_1} = (3,\,2,\,-6)$$.

Line 2: $$6x = -y = -4z$$
Put the common value as $$\mu$$: $$6x = -y = -4z = \mu$$.
Hence $$x = \dfrac{\mu}{6},\; y = -\mu,\; z = -\dfrac{\mu}{4}$$.
So the direction ratios are proportional to $$\left(\dfrac{1}{6},\,-1,\,-\dfrac{1}{4}\right)$$.
Multiply by $$12$$ to clear denominators: second line d.r.’s $$\mathbf{a_2} = (2,\,-12,\,-3)$$.

The angle $$\theta$$ between two lines with direction ratios $$\mathbf{a_1} = (l_1,m_1,n_1)$$ and $$\mathbf{a_2} = (l_2,m_2,n_2)$$ is given by
$$\cos\theta = \dfrac{l_1l_2 + m_1m_2 + n_1n_2}{\sqrt{l_1^{2}+m_1^{2}+n_1^{2}}\;\sqrt{l_2^{2}+m_2^{2}+n_2^{2}}}\,.$$

Compute the numerator (dot product):
$$l_1l_2 + m_1m_2 + n_1n_2 = 3\cdot2 + 2\cdot(-12) + (-6)\cdot(-3) = 6 - 24 + 18 = 0.$$

Because the dot product is $$0$$, $$\cos\theta = 0$$, which implies $$\theta = 90^\circ$$.

Hence the two lines are perpendicular.

Option B which is: $$90^\circ$$