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 = $${1 \over 2} \times \left| {\overline {BC} } \right| \times \left| {\overline {AD} } \right|$$

= $${1 \over 2} \times 5 \times {{2\sqrt {34} } \over 5} = \sqrt {34} $$