Let a line L be perpendicular to both the lines L_1: x+1 3 = y+3 5 = z+5 7 and L_2: x-2 1 = y-4 4 = z-6 7. If θ is the acute angle between the lines…

The direction vector of line $L_1$L1​ is $\vec{d_1} = 3\hat{i} + 5\hat{j} + 7\hat{k}$d1​​=3i^+5j^​+7k^.

The direction vector of line $L_2$L2​ is $\vec{d_2} = \hat{i} + 4\hat{j} + 7\hat{k}$d2​​=i^+4j^​+7k^.

Since line $L$L is perpendicular to both $L_1$L1​ and $L_2$L2​, its direction vector $\vec{d}$d is given by the cross product of $\vec{d_1}$d1​​ and $\vec{d_2}$d2​​:

$$\vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{vmatrix}$$d=d1​​×d2​​=​i^31​j^​54​k^77​​

$$\vec{d} = \hat{i}(35 - 28) - \hat{j}(21 - 7) + \hat{k}(12 - 5) = 7\hat{i} - 14\hat{j} + 7\hat{k}$$d=i^(35−28)−j^​(21−7)+k^(12−5)=7i^−14j^​+7k^

The direction ratios of $L$L are proportional to $1, -2, 1$1,−2,1.

The direction vector of line $L_3$L3​ is $\vec{d_3} = 2\hat{i} + \hat{j} + 2\hat{k}$d3​​=2i^+j^​+2k^.

The acute angle $\theta$θ between $L$L and $L_3$L3​ is given by:

$$\cos\theta = \dfrac{|\vec{d} \cdot \vec{d_3}|}{|\vec{d}| |\vec{d_3}|}$$cosθ=∣d∣∣d3​​∣∣d⋅d3​​∣​

$$\cos\theta = \dfrac{|(1)(2) + (-2)(1) + (1)(2)|}{\sqrt{1^2 + (-2)^2 + 1^2} \sqrt{2^2 + 1^2 + 2^2}}$$cosθ=12+(−2)2+12​22+12+22​∣(1)(2)+(−2)(1)+(1)(2)∣​

$$\cos\theta = \dfrac{|2 - 2 + 2|}{\sqrt{6} \sqrt{9}} = \dfrac{2}{3\sqrt{6}}$$cosθ=6​9​∣2−2+2∣​=36​2​

From $\cos\theta = \dfrac{2}{3\sqrt{6}}$cosθ=36​2​, considering a right-angled triangle with adjacent side $2$2 and hypotenuse $3\sqrt{6}$36​:

The opposite side is $\sqrt{(3\sqrt{6})^2 - 2^2} = \sqrt{54 - 4} = \sqrt{50} = 5\sqrt{2}$(36​)2−22​=54−4​=50​=52​.

Therefore, $\tan\theta = \dfrac{5\sqrt{2}}{2}$tanθ=252​​.

Answer: $\dfrac{5}{2}\sqrt{2}$25​2​