JEE Advance - Mathematics (2005 - No. 3)
Find the equation of the plane containing the line $$2x-y+z-3=0,3x+y+z=5$$ and at a distance of $${1 \over {\sqrt 6 }}$$ from the point $$(2, 1, -1).$$
$$62x+29y+19z-105=0$$ and $$2x-y+z-3=0$$
$$5x + 2y + 2z = 0$$ and $$x + y + z = 1$$
$$62x+29y+19z-105=0$$ and $$50x+31y+21z-135=0$$
$$x+y+z = 0$$ and $$x-y-z=1$$
$$x=0, y=0, z=0$$
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