pls answer question 27

Dear Student,
Please find below the solution to the asked query:

We haveAx1,y1,z1=0,5,0Bx2,y2,z2=0,0,12Cx3,y3,z3=0,5,12c=AB=02+52+122=13a=BC=02+52+02=5b=AC=02+02+122=12Incentre is ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c,az1+bz2+cz3a+b+c=5×0+12×0+13×05+12+13,5×5+12×0+13×55+12+13,5×0+12×12+13×125+12+13=0,9030,30030=0,3,10

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