If the angle between the line $$x = \frac{{y - 1}}{2} = \frac{{z - 3}}{\lambda }$$    and the plane $$x+2y+3z=4$$    is $${\cos ^{ - 1}}\left( {\sqrt {\frac{5}{{14}}} } \right),$$    then $$\lambda $$ equals :

Solution :
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 $$\theta = {\cos ^{ - 1}}\sqrt {\frac{5}{{14}}} = \sin \,\theta = \frac{3}{{\sqrt {14} }}$$
$$\therefore \frac{{5 + 3\lambda }}{{\sqrt {14} .\sqrt {5 + {\lambda ^2}} }} = \frac{3}{{\sqrt {14} }}\,\,\,\,\,\,\,\,\,\, \Rightarrow \lambda = \frac{2}{3}$$