The angle between the lines $$3x-3y=-2z$$ and $$2x-y = -3z$$​ is:

Correct option is A

​$$\begin{aligned}&\text{1. Convert to symmetric form:} \\&\frac{x}{\frac{1}{3}} = \frac{y}{\frac{1}{3}} = \frac{z}{-\frac{1}{2}} \quad \text{and} \quad \frac{x}{\frac{1}{2}} = \frac{y}{-1} = \frac{z}{-\frac{1}{3}} \\&\text{2. Identify direction vectors:} \\&\vec{a} = \left(\frac{1}{3}, \frac{1}{3}, -\frac{1}{2} \right), \quad \vec{b} = \left(\frac{1}{2}, -1, -\frac{1}{3} \right) \\&\text{3. Compute dot product:} \\&\vec{a} \cdot \vec{b} = \left(\frac{1}{3} \times \frac{1}{2} \right) + \left(\frac{1}{3} \times -1\right) + \left(-\frac{1}{2} \times -\frac{1}{3} \right) = \frac{1}{6} - \frac{1}{3} + \frac{1}{6} = 0 \\&\text{4. Calculate magnitudes:} \\&|\vec{a}| = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{1}{9} + \frac{1}{9} + \frac{1}{4}} = \sqrt{\frac{17}{36}} = \frac{\sqrt{17}}{6} \\&|\vec{b}| = \sqrt{\left(\frac{1}{2}\right)^2 + (-1)^2 + \left(-\frac{1}{3}\right)^2} = \sqrt{\frac{1}{4} + 1 + \frac{1}{9}} = \sqrt{\frac{49}{36}} = \frac{7}{6} \\&\text{5. Find angle:} \\&\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{0}{\frac{\sqrt{17}}{6} \times \frac{7}{6}} = 0 \\&\theta = \cos^{-1}(0) = 90^\circ \\&\text{Final Answer:} \\&90^\circ\end{aligned}$$​​