If the angle between the lines whose direction ratios are (2,-1,2)...

Given direction ratios are (2,-1,2)$and (\mathrm{x}, 3,5)$ We know that the angle between the lines whose direction ratios are \left(a_1, b_1, c_1\right)$and \left(a_2, b_2, c_2\right)$ is $$\begin{aligned} & \cos \theta=\frac{\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{ b}_2+\mathrm{c}_1 \mathrm{c}_2}{\sqrt{\mathrm{a}_1^2+\mathrm{b}_1^2+\mathrm{c}_1^2} \sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2+\mathrm{c}_2^2}} \\ & \Rightarrow \cos \frac{\pi}{4}=\frac{2 \mathrm{x}-3+10}{\sqrt{4+1+4} \sqrt{\mathrm{x}^2+9+25}}=\frac{2 \mathrm{x}+7}{\sqrt{9} \cdot \sqrt{\mathrm{x}^2+34}}\end{aligned}$$\Rightarrow \frac{1}{\sqrt{2}}=\frac{2 x+7}{3 \sqrt{x^2+34}} \Rightarrow 2 x+7=3 \sqrt{\frac{x^2+34}{2}}\Rightarrow 4 \mathrm{x}^2+49+28 \mathrm{x}=\frac{9\left(\mathrm{x}^2+34\right)}${2}$ (Squaring on both sides) $$\begin{aligned} & \Rightarrow 2\left(4 x^2+49+28 x\right)=9 x^2+306 \\ & \Rightarrow 8 x^2+98+56 x=9 x^2+306 \\ & \Rightarrow x^2-56 x+208=0\end{aligned}$$$$\begin{aligned} & \therefore \mathrm{x}=\frac{56 \pm \sqrt{3136-812}}{2}=\frac{56 \pm 48}{2}=28 \pm 24 \\ & =4,52 \\ & \text{Smaller value }=4\end{aligned}$$