Answer:

Direction ratios of the x-axis are 1,0,0 and direction ratios of normal to the plane are 2,−3,−6.

The angle between the line and the plane,

\sin \phi=\frac|{\vec{a} \cdot \vec{n} \mid}{\vec{a}|| \vec{n} \mid}

$\sin \varphi =\frac{|}{a_1,a_2+b_1·b_2+c_1·c_2\mid }\sqrt{a_1^2+b_1^2–c_1^2·\sqrt{a_2^2–b_2^2+c_2^2}}$

\sin \phi=\frac|{a_{1}, a_{2}+b_{1} \cdot b_{2}+c_{1} \cdot c_{2} \mid}{\sqrt{a_{1}^{2}+b_{1}^{2}-c_{1}^{2} \cdot \sqrt{a_{2}^{2}-b_{2}^{2}+c_{2}^{2}}}}

Let be the angle between the -axis and the given plane.

\begin{array}{l}

\sin \phi=\frac{|1 \times 2+0 \times(-3)+0 \times(6)|}{\left\{\sqrt{\left.1^{2}+0^{2}+0^{2}\right\}}\left\{\sqrt{2^{2}+(-3)^{2}+(-6)^{2}}\right\}\right.}=\frac{2}{7} \\

\Rightarrow \phi=\sin ^{1}\left(\frac{2}{7}\right)

\end{array}

Hence, .