What is the angle between the line \(6 x=4 y=3 z\) and the plane \(3 x+2 y-3 z=4\) ?

The equation of the given line is \(6 x=4 y=3 z\) which is written in symmetric form as \(\frac{x-0}{1 / 6}=\frac{y-0}{1 / 4}=\frac{z-0}{1 / 3}\) Direction ratios of this line are \(\left(\frac{1}{6}, \frac{1}{4}, \frac{1}{3}\right)\) and equation of the plane is \(3 x+2 y-3 z-4=0\) If \(\theta\) be the angle between line and plane, then direction ratios of the normal to this plane is (3, \(2,-3\) ) \(\begin{aligned} \sin \theta & =\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right| \\ & =\left|\frac{\frac{1}{6} \times 3+\frac{1}{4} \times 2+\frac{1}{3}(-3)}{\sqrt{\frac{1}{36}+\frac{1}{16}+\frac{1}{9}} \sqrt{9+4+9}}\right|=0 \\ \Rightarrow \quad \theta & =0^{\circ} \end{aligned}\)