Substitute in (i) we get \(\frac { x }{ 3\alpha } +\frac { y }{ 3\beta } +\frac { z }{ 3\gamma } =1\Rightarrow \frac { x }{ \alpha } +\frac { y }{ \beta } +\frac { z }{ \gamma } =3\)

Since \(l, m, n \text { and } l+\delta l, m+\delta m, n+\delta n\) are direction cosines of a variable line in two different positions,therefore,
\(l^{2}+m^{2}+n^{2}=1\)
\(\text { and }(l+\delta l)^{2}+(m+\delta m)^{2}+(n+\delta n)^{2}=1\)
\(\text { Now, }(l+\delta l)^{2}+(m+\delta m)^{2}+(n+\delta n)^{2}=1\)
\(\Rightarrow\left(l^{2}+m^{2}+n^{2}\right)+2(l \cdot \delta l+m \cdot \delta m+n \cdot \delta n) \)
\(+(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}=1[\text { from }(i i)] \)
\(\Rightarrow 1+2(l \delta l+m \delta m+n \delta n)+(\delta l)^{2} \)
\(+(\delta m)^{2}+(\delta n)^{2}=1[\text { using }(i)] \)
\(\Rightarrow 2(l \delta l+m \delta m+n \delta n)=-\left[(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}\right] \)
\(\Rightarrow(l \delta l+m \delta m+n \delta n)=-\frac{1}{2}\left[(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}\right] \)
\(\text { Also } \cos (\delta \theta)=l(l+\delta l)+m(m+\delta m)+n(n+\delta) \)
\(=\left(l^{2}+m^{2}+n^{2}\right)+(l \delta l+m \delta m+n \delta n) \)
\(=1-\frac{1}{2}\left[(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}\right] \quad[\text { using }(i) \text { and (iii)] }\)
\(\Rightarrow 2(1-\cos \delta \theta)=(\delta I)^{2}+(\delta m)^{2}+(\delta n)^{2}\)
\(\Rightarrow 2 \times 2 \sin ^{2} \frac{\delta \theta}{2}=(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}\)
\(\left[\because 1-\cos 2 \theta=2 \sin ^{2} \theta\right]\)
\(\Rightarrow 4\left[\frac{\delta \theta}{2}\right]^{2}=(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}\)
\(\text { [if } \theta \text { is small, then } \sin \theta \rightarrow \theta]\)
\(\Rightarrow(\delta \theta)^{2}=(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2} \)
which is true.

If lines intersect then \(\left(\vec{a}_{2}-\overrightarrow{a_{1}}\right) \cdot\left(\vec{b}_{1} \times \overrightarrow{b_{2}}\right)=0\)
\(\text { i.e. }\left|\begin{array}{ccc} 4-1 & 1-2 & 0-3 \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{array}\right|=0 \)
\(\Rightarrow 3(3-8)+1(2-20)-3(4-15)=0 \)
\(\Rightarrow-15-18+33=0, \text { true }\)
Hence, the lines intersect
For point of intersection, take general points on the two
\(\text {lines as }(2 \lambda+1,3 \lambda+2,4 \lambda+3)\text { and }(5 \mu+4,2 \mu+1, \mu)\)
If lines intersect then for same values \(\lambda \text { and } \mu \text {. }\)
\(2 \lambda+1=5 \mu+4 \Rightarrow 2 \lambda-5 \mu=3 \)
\(3 \lambda+2=2 \mu+1 \Rightarrow 3 \lambda-2 \mu=-1 \)
\(4 \lambda+3=\mu \Rightarrow 4 \lambda-\mu=-3\)
Solving \((i i) \text { and (iii) for } \lambda, \mu \text {, we get }\)
\(\lambda=-1, \mu=-1\)
Substituting in (iv), we get -4+1 = -3 true
Hence, for \(\lambda=-1, \mu=-1 \) the lines intersect.
Substituting in \( (i) \text {, for } \lambda=-1 \text {, }\)
Point of intersection is \( (-2+1,-3+2,-4+3) \text {, i.e }(-1,-1,-1) \)

The equation of a plane through the line of intersection of the planes
\(x+2 y+3 z-4=0 \text { and } 2 x+y-z+5=0\) is 
\((x+2 y+3 z+4)+\lambda(2 x+y-z+5)=0\) 
As, this is perpendicular to the plane
\(5 x+3 y+6 z+8=0\) 
\(\therefore \quad 5(1+2 \lambda)+3(2+\lambda)+6(3-\lambda)=0\) 
\(\left[\because a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\right]\) 
\(\Rightarrow 5+10 \lambda+6+3 \lambda+18-6 \lambda=0\) 
\(\lambda=-\frac{29}{7}\) 
Then, from Eq. (i), we get
\(x\left[1+2\left(\frac{-29}{7}\right)\right]+y\left(2-\frac{29}{7}\right)+z\left(\frac{29}{7}+3\right)-4+5\left(\frac{-29}{7}\right)=0\) 
\(\Rightarrow x(7-58)+y(14-29)+z(29+21)-28-145=0\) 
\(\Rightarrow -51 x-15 y+50 z-173=0\) 
Hence, the required equation of plane is
\(51 x+15 y-50 z+173=0\)