Official solution
Area of \[\Delta \]ABC = \[{1 \over 2} \times BC \times AD\]
Given BC = 5 so we need perpendicular
distance of A from line BC.
Let a point D on BC = (3\[\lambda \] - 2, 1, 4\[\lambda \])
Direction ratio of AD
3\[\lambda \] - 3, 2, 4\[\lambda \] - 2
As AD perpendicular to the BC, so
DR of AD. DR of BC = 0
\[(3\lambda - 3)3 + 2(0) + (4\lambda - 2)4 = 0\]
\[9\lambda - 9 + 16\lambda - 8 = 0 \Rightarrow \lambda = {{17} \over {25}}\]
Hence, \[D = \left( {{1 \over {25}},1,{{68} \over {25}}} \right)\]
\[\left| {\overline {AD} } \right| = \sqrt {{{\left( {{1 \over {25}} - 1} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( {{{68} \over {25}} - 2} \right)}^2}} \]
\[ \Rightarrow \sqrt {{{\left( {{{ - 24} \over {25}}} \right)}^2} + 4 + {{\left( {{{18} \over {25}}} \right)}^2}} \]
\[ \Rightarrow \sqrt {{{{{\left( {24} \right)}^2} + 4{{\left( {25} \right)}^2} + {{\left( {18} \right)}^2}} \over {{{25}^2}}}} \]
\[ \Rightarrow \sqrt {{{576 + 2500 + 324} \over {{{25}^2}}}} \]
\[ \Rightarrow \sqrt {{{3400} \over {{{25}^2}}}} \Rightarrow {{\sqrt {34} .10} \over {25}} = {{2\sqrt {34} } \over 5}\]
Area of triangle = \[{1 \over 2} \times \left| {\overline {BC} } \right| \times \left| {\overline {AD} } \right|\]
= \[{1 \over 2} \times 5 \times {{2\sqrt {34} } \over 5} = \sqrt {34} \]