Find the area of the triangle having vertices at X132 Y2 11

Find the area of the triangle having vertices at X(1,3,2), Y(2, -1,1) and Z(-1,2,3)

Express this problem by using the three dimensional distance formula.

area of triangle = 1/2*IaxbI

a is vector from X(1,3,2) Y(2, -1,1) = (1 , -4 , -1)

b is vector from X(1,3,2) Z(-1,2,3) = ( -2 , -1 , 1)

=> axb = {-5; 1; -9}

=> area = 1/2*sqrt(5^2+1+9^2) = 5.17 unit