If the angle between the lines x 2= y 2= z 1 and 5-x -2= 7y-14 P= z-3 4 is cos^-1( 2 3), then P is equal to

Given lines are $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$2x​=2y​=1z​ and $\frac{5-x}{-2}=\frac{7y-14}{p}=\frac{z-3}{4}$−25−x​=p7y−14​=4z−3​

$$\Rightarrow \frac{x-5}{2}=\frac{y-2}{(\frac{P}{7})}=\frac{z-3}{4}$$⇒2x−5​=(7P​)y−2​=4z−3​

We know that the angle between the lines $\frac{x-{x}_{1}}{{a}_{1}}=\frac{y-{y}_{1}}{{b}_{1}}=\frac{z-{z}_{1}}{{c}_{1}}$a1​x−x1​​=b1​y−y1​​=c1​z−z1​​ and $\frac{x-{x}_{2}}{{a}_{2}}=\frac{y-{y}_{2}}{{b}_{2}}=\frac{z-{z}_{2}}{{c}_{2}}$a2​x−x2​​=b2​y−y2​​=c2​z−z2​​ is given by $\theta ={\mathrm{cos}}^{-1}|\frac{{a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}+{c}_{1}^{2}}\times \sqrt{{a}_{2}^{2}+{b}_{2}^{2}+{c}_{2}^{2}}}|$θ=cos−1∣a12​+b12​+c12​​×a22​+b22​+c22​​a1​a2​+b1​b2​+c1​c2​​∣

Given, the angle between both lines is ${\mathrm{cos}}^{-1}(\frac{2}{3})$cos−1(32​)

$$\Rightarrow {\mathrm{cos}}^{-1}(\frac{2}{3})={\mathrm{cos}}^{-1}(\frac{2\times 2+2\times \frac{P}{7}+1\times 4}{\sqrt{{2}^{2}+{2}^{2}+{1}^{2}}\times \sqrt{{2}^{2}+{(\frac{P}{7})}^{2}+{4}^{2}}})$$⇒cos−1(32​)=cos−1(22+22+12​×22+(7P​)2+42​2×2+2×7P​+1×4​)

$$\Rightarrow \frac{2}{3}=\frac{4+\frac{2P}{7}+4}{3\times \sqrt{4+\frac{{P}^{2}}{49}+16}}$$⇒32​=3×4+49P2​+16​4+72P​+4​

$$\Rightarrow \frac{2}{3}=\frac{56+2P}{3\times \sqrt{{P}^{2}+980}}$$⇒32​=3×P2+980​56+2P​

$$\Rightarrow \sqrt{{P}^{2}+980}=P+28$$⇒P2+980​=P+28

$$\Rightarrow {P}^{2}+980={(P+28)}^{2}$$⇒P2+980=(P+28)2

$$\Rightarrow {P}^{2}+980={P}^{2}+56P+784$$⇒P2+980=P2+56P+784

$$\Rightarrow 56P=196$$⇒56P=196

$$\Rightarrow P=\frac{7}{2}.$$⇒P=27​.