It is given that, |a|=1,|b|=2,|c|=3,b.c=0 Then, |2a+3b−3c|=√4|a|2+9|b|2+12a.b−18b.c−12a.c =√4+9(4)+9(9)+12a.b−0−12a.c =√4+36+81+12a.b−12a.c The projection of b along a is equal to the projection of c along a so, a.b = a.c Thus, |2a+3b−3c|=√121+12a.b−12a.b =11