1.Let the distance between the planes 3x+4y+5z+c1=03x+4y+5z+c_1=03x+4y+5z+c1=0 and 3x+4y+5z−c2=03x+4y+5z-c_2=03x+4y+5z−c2=0 be 222\sqrt{2}22, where c1,c2∈Wc_1,c_2\in\mathbb{W}c1,c2∈W (whole numbers). Then the number of possible ordered pairs (c1,c2)(c_1,c_2)(c1,c2) isa.111b.212121c.313131d.515151Login to continueOnly logged in users canattempt or see the solution.