$$\frac{{x - b}}{a} = \frac{y}{1} = \frac{{z - d}}{c}\,;\,\frac{{x - b'}}{{a'}} = \frac{y}{1} = \frac{{z - d'}}{{c'}}$$
For perpendicularity of lines $$aa'+1+cc'=0$$

Given two lines are :
$$\eqalign{ & y = \frac{m}{\ell }x + \alpha ,\,z = \frac{n}{\ell }x + \beta \,{\text{ and}} \cr & y = \frac{{m'}}{{\ell '}}x + \alpha ',\,z = \frac{{n'}}{{\ell '}}x + \beta ' \cr} $$
These two lines can be represented as :
$$\eqalign{ & \frac{{y - \alpha }}{{\frac{m}{\ell }}} = \frac{{x - 0}}{1} = \frac{{z - \beta }}{{\frac{n}{\ell }}} \cr & {\text{and }}\frac{{y - \alpha '}}{{\frac{{m'}}{{c'}}}} = \frac{{x - 0}}{1} = \frac{{z - \beta '}}{{\frac{{n'}}{{\ell '}}}} \cr} $$
They are orthogonal, if
$$\frac{m}{\ell } \times \frac{{m'}}{{\ell '}} + 1 \times 1 + \frac{n}{\ell } \times \frac{{n'}}{{\ell '}} = - 1 \Rightarrow \ell \ell ' + mm' + nn' = 0$$

Perpendicular distance of centre $$\left( {\frac{1}{2},\,0,\, - \frac{1}{2}} \right)$$   from $$x+2y-2=4$$    is given by $$\frac{{\left| {\frac{1}{2} + \frac{1}{2} - 4} \right|}}{{\sqrt 6 }} = \sqrt {\frac{3}{2}} $$
Radius of sphere $$ = \sqrt {\frac{1}{4} + \frac{1}{4} + 2} = \sqrt {\frac{5}{2}} $$
$$\therefore $$ Radius of circle $$ = \sqrt {\frac{5}{2} - \frac{3}{2}} = 1$$

If $$\theta $$ be the angle between the given line and plane, then
$$\sin \,\theta = \frac{{1 \times 1 + 2 \times 2 + \lambda \times 3}}{{\sqrt {{1^2} + {2^2} + {\lambda ^2}} .\sqrt {{1^2} + {2^2} + {3^2}} }} = \frac{{5 + 3\lambda }}{{\sqrt {14} .\sqrt {5 + {\lambda ^2}} }}$$
But it is given that $$\theta = {\cos ^{ - 1}}\sqrt {\frac{5}{{14}}} = \sin \,\theta = \frac{3}{{\sqrt {14} }}$$
$$\therefore \frac{{5 + 3\lambda }}{{\sqrt {14} .\sqrt {5 + {\lambda ^2}} }} = \frac{3}{{\sqrt {14} }}\,\,\,\,\,\,\,\,\,\, \Rightarrow \lambda = \frac{2}{3}$$

Let the points are $$A,\,B,\,C$$   and $$D$$ respectively.
Mid point of $$AC$$  is
$$\left( {\frac{{4 - 1}}{2},\,\frac{{7 - 2}}{2},\,\frac{{8 + 1}}{2}} \right){\text{ or }}\left( {\frac{3}{2},\,\frac{5}{2},\,\frac{9}{2}} \right)$$
Mid point of $$BD$$  is
$$\left( {\frac{{2 + 1}}{2},\,\frac{{3 + 2}}{2},\,\frac{{4 + 5}}{2}} \right){\text{ or }}\left( {\frac{3}{2},\,\frac{5}{2},\,\frac{9}{2}} \right)$$
Thus $$AC$$  and $$BD$$  bisect each other. Further,
$$\eqalign{ & AC = \sqrt {{{\left( {4 + 1} \right)}^2} + {{\left( {7 + 2} \right)}^2} + {{\left( {8 - 1} \right)}^2}} \cr & AC = \sqrt {25 + 81 + 49} \cr & AC = \sqrt {155} \cr & BD = \sqrt {{{\left( {2 - 1} \right)}^2} + {{\left( {3 - 2} \right)}^2} + {{\left( {4 - 5} \right)}^2}} \cr & BD = \sqrt {1 + 1 + 1} \cr & BD = \sqrt 3 \cr & \therefore \,AC \ne BD \cr} $$
Hence, $$ABCD$$   represents a parallelogram.

Let the plane be $$a\left( {x - 1} \right) + b\left( {y + 1} \right) + c\left( {z + 1} \right) = 0$$
Normal vector
\[\left| \begin{array}{l} \hat i\,\,\,\,\,\,\,\,\,\,\,\hat j\,\,\,\,\,\,\,\,\,\,\,\,\hat k\\ 1\,\,\, - 2\,\,\,\,\,\,\,\,\,\,3\\ 2\,\,\, - 1\,\, - 1 \end{array} \right| = 5\hat i + 7\hat j + 3\hat k\]
So plane is
$$\eqalign{ & 5\left( {x - 1} \right) + 7\left( {y + 1} \right) + 3\left( {z + 1} \right) = 0 \cr & \Rightarrow 5x + 7y + 3z + 5 = 0 \cr} $$
Distance of point $$\left( {1,\,3,\, - 7} \right)$$   from the plane is
$$\frac{{5 + 21 - 21 + 5}}{{\sqrt {25 + 49 + 9} }} = \frac{{10}}{{\sqrt {83} }}$$

The equation of the line is $$\frac{{x - 2}}{2} = \frac{{y - 3}}{3} = \frac{{z - 4}}{5} = r$$
Where $$r$$ is a constant. Any point on this line, is given by $$x = 2r + 2,\,y = 3r + 2$$     and $$z = 5r + 4$$
Since, a plane that is parallel to $$z$$-axis will have no $$z$$-co-ordinate, $$z = 0$$
$$z = 0 \Rightarrow 5r + 4 = 0{\text{ or, }}r = \frac{{ - 4}}{5}$$
Putting this value of r for $$x$$ and $$y$$ co-ordinates.
$$\eqalign{ & x = 2r + 2 = 2 \times \left( { - \frac{4}{5}} \right) + 2 \cr & {\text{or, }}5x = - 8 + 10 \cr & {\text{or, }}5x = 2 \cr & x = \frac{2}{5},\,{\text{or }}\,\frac{2}{x} = 5......\left( 1 \right) \cr & {\text{Similarly, }}y = 3r + 3 = 3 \times \left( { - \frac{4}{5}} \right) + 3 \cr & {\text{or, }}5y = - 12 + 15 \cr & {\text{or, }}5y = 3 \cr & y = \frac{3}{5},\,{\text{or }}\frac{3}{y} = 5......\left( 2 \right) \cr & {\text{From equation}}\,\left( 1 \right){\text{ and }}\left( 2 \right) \cr & \frac{2}{x} = \frac{3}{y} \Rightarrow 3x - 2y = 0 \cr} $$

$$\eqalign{ & {\text{Let the equation of the plane be}} \cr & \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \cr & \Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 \cr & \Rightarrow {\text{Volume of tetrahedron}}\,OABC = V = \frac{1}{6}\left( {abc} \right) \cr & {\text{Now, }}{\left( {abc} \right)^{\frac{1}{3}}} \geqslant \frac{3}{{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}} \geqslant 3\left( {{\text{G}}{\text{.M}}{\text{.}} \geqslant {\text{H}}{\text{.M}}{\text{.}}} \right) \cr & {\text{or, }}abc \geqslant 27 \Rightarrow V \geqslant \frac{9}{2} \cr} $$

$$\eqalign{ & \because \,\overrightarrow a {\text{ lies in the plane of }}\overrightarrow b {\text{ and }}\overrightarrow c \cr & \therefore \,\overrightarrow a = \overrightarrow b + \lambda \overrightarrow c \cr & \Rightarrow \alpha \hat i + 2\hat j + \beta \hat k = \hat i + \hat j + \lambda \left( {\hat j + \hat k} \right) \cr & \Rightarrow \alpha = 1,\,2 = 1 + \lambda ,\,\beta = \lambda \cr & \Rightarrow \alpha = 1,\,\,\beta = 1 \cr} $$

$$\eqalign{ & {\text{Here, }}\alpha = \beta = \gamma \cr & \because \,{\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1 \cr & \therefore \,\cos \,\alpha = \frac{1}{{\sqrt 3 }} \cr & {\text{Direction cosines of }}PQ{\text{ are }}\left( {\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }}} \right) \cr} $$

$$\eqalign{ & PM = {\text{Projection of }}AP{\text{ on }}PQ \cr & = \left| {\left( { - 2 + 3} \right)\frac{1}{{\sqrt 3 }} + \left( {3 - 5} \right)\frac{1}{{\sqrt 3 }} + \left( {1 - 2} \right)\frac{1}{{\sqrt 3 }}} \right| \cr & = \frac{2}{{\sqrt 3 }} \cr & {\text{and}}\,AP = \sqrt {{{\left( { - 2 + 3} \right)}^2} + {{\left( {3 - 5} \right)}^2} + {{\left( {1 - 2} \right)}^2}} = \sqrt 6 \cr & AM = \sqrt {{{\left( {AP} \right)}^2} - {{\left( {PM} \right)}^2}} = \sqrt {6 - \frac{4}{3}} = \sqrt {\frac{{14}}{3}} \cr} $$