According to the given ques:

In \[\vartriangle ABC,\text{ }a\text{ }=\text{ }\surd 2,\text{ }b\text{ }=\text{ }\surd 3\text{ }and\text{ }c\text{ }=\text{ }\surd 5~\]

Hence, using the formulas,

\[cos\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})/2bc\]

Substituting the values in the equation, we get,

\[=\text{ }[{{\left( \surd 3 \right)}^{2}}~+\text{ }{{\left( \surd 5 \right)}^{2}}~\text{ }{{\left( \surd 2 \right)}^{2}}\left] \text{ }/\text{ } \right[2\text{ }\times \text{ }\surd 3\text{ }\times \text{ }\surd 5]\]

\[=\text{ }3/\surd 15\]

Since, Area of \[\vartriangle ABC\text{ }=\text{ }1/2\text{ }bc\text{ }sin\text{ }A\]

Finding sin A:

\[Sin\text{ }A\text{ }=\text{ }\surd (1\text{ }\text{ }co{{s}^{2}}~A)\]

\[\left[ by\text{ }using\text{ }trigonometric\text{ }identity \right]\]

Or,

\[=\text{ }\surd (1\text{ }\text{ }{{\left( 3/\surd 15 \right)}^{2}})\]

\[=\text{ }\surd \left( 1-\text{ }\left( 9/15 \right) \right)\]

So,

\[=\text{ }\surd \left( 6/15 \right)\]

Now,

\[Area\text{ }of\text{ }\vartriangle ABC\text{ }=\text{ }1/2\text{ }bc\text{ }sin\text{ }A\]

\[=\text{ }1/2\text{ }\times \text{ }\surd 3\text{ }\times \text{ }\surd 5\text{ }\times \text{ }\surd \left( 6/15 \right)\]

\[=\text{ }1/2\text{ }\surd 6\text{ }sq.\text{ }units\]

Hence proved.