Find the area of the triangle with vertices A(1,1,2), B(2,3,5) and C(1,5,5).

√(61)/2

The vertices of the triangle ABC is given  as A(1,1,2), B(2,3,5) and C(1,5,5)

The adjacent sides \(\vec{AB}\)  and \(\vec{BC}\)  of triangle ABC are given as:

\(\vec{AB}\) =  (2-1)\(\hat{i}\) + (3-1)\(\hat{j}\) + (5-2) \(\hat{k}\)  =\(\hat{i}\) + 2\(\hat{j}\) + 3 \(\hat{k}\)

\(\vec{BC}\)=  (1-2)\(\hat{i}\) + (5-3)\(\hat{j}\)+ (5-5)\(\hat{k}\)  = -\(\hat{i}\) + 2\(\hat{j}\)

Area of the  triangle ABC  = 1/2 {|\(\vec{AB}\)x\(\vec{BC}\)|} 

\(\vec{AB}\)x\(\vec{BC}\) =( \(\hat{i}\) + 2\(\hat{j}\)+ 3 \(\hat{k}\))x (-\(\hat{i}\)+ 2\(\hat{j}\) + 0\(\hat{k}\))

                = (-6) \(\hat{i}\) -  3\(\hat{j}\) + (2+2)\(\hat{k}\)

                = -6 \(\hat{i}\) -  3\(\hat{j}\)+ 4\(\hat{k}\)

so, |\(\vec{AB}\)x\(\vec{BC}\)| = √(-6)²+ (-3)² +(4)² = √61

Hence, Area of the  triangle ABC  =  √(61)/2