1.If a⃗×b⃗×c⃗=0\vec{a} \times \vec{b} \times \vec{c} = 0a×b×c=0, a⃗×b⃗+b⃗×c⃗+c⃗×a⃗=6(b⃗×c⃗)\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = 6(\vec{b} \times \vec{c})a×b+b×c+c×a=6(b×c), then the locus of the point (x,y)(x, y)(x,y) isa.x2+y2=1x^2 + y^2 = 1x2+y2=1b.xy=5xy = 5xy=5c.2x+6y=52x + 6y = 52x+6y=5d.x+y+6=0x + y + 6 = 0x+y+6=0Login to continueOnly logged in users canattempt or see the solution.
