let L be line of intersection of planes 2x+3y+z=1 and x+3y+2z=2.if L makes an angleα
with positive x axis then prove that cos α =1/√3

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ 2\mathrm{x}+3\mathrm{y}+\mathrm{z}=1 \\ \Rightarrow 2\mathrm{x}+3\mathrm{y}=1-\mathrm{z} ;\left(\mathrm{i}\right) \\ \mathrm{x}+3\mathrm{y}+2\mathrm{z}=2 \\ \Rightarrow \mathrm{x}+3\mathrm{y}=2-2\mathrm{z} ;\left(\mathrm{ii}\right) \\ \left(\mathrm{i}\right)-\left(\mathrm{ii}\right) \\ \Rightarrow 2\mathrm{x}+3\mathrm{y}-\mathrm{x}-3\mathrm{y}=1-\mathrm{z}-2+2\mathrm{z} \\ \Rightarrow \mathrm{x}=\mathrm{z}-1 \\ \Rightarrow \boxed{\mathrm{z}=\frac{\mathrm{x}+1}{1}} \\ \mathrm{Putting} \mathrm{value} \mathrm{of} \mathrm{z} \mathrm{in} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get}, \\ \mathrm{z}-1+3\mathrm{y}=2-2\mathrm{z} \\ \Rightarrow 3\mathrm{y}=2-2\mathrm{z}-\mathrm{z}+1 \\ \Rightarrow 3\mathrm{y}=-3\left(\mathrm{z}-1\right) \\ \Rightarrow \mathrm{y}=-\left(\mathrm{z}-1\right)=-\mathrm{z}+1 \\ \Rightarrow \boxed{\frac{\mathrm{y}-1}{-1}=\mathrm{z}} \\ \mathrm{Hence}, \mathrm{we} \mathrm{have} \\ \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}-1}{-1}=\frac{\mathrm{z}}{1} \\ \mathrm{For} \frac{\mathrm{x}-{\mathrm{x}}_1}{\mathrm{a}}=\frac{\mathrm{y}-{\mathrm{y}}_1}{\mathrm{b}}=\frac{\mathrm{z}-{\mathrm{z}}_1}{\mathrm{c}}, \mathrm{cos\alpha } \mathrm{is} \mathrm{given} \mathrm{by}: \\ \mathrm{cos\alpha }=\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}} \\ \mathrm{Hence} \mathrm{for} \mathrm{given} \mathrm{question}: \\ \mathrm{cos\alpha }=\frac{1}{\sqrt{1^2+{\left(-1\right)}^2+1^2}} \\ \Rightarrow \boxed{\mathbf{cos\alpha }\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{3}}}} \left(\mathrm{Hence} \mathrm{proved}\right) \end{gathered}$$

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