Let A (1$$, $$1$$, $$1), B (2$$,$$3$$,$$5) and C $$(-1$$ ,$$0$$,$$2)$$ be three points, then equation of a plane parallel to the plane ABC which is at distance $$2$$ is

Question

Let A (1$$, $$1$$, $$1), B (2$$,$$3$$,$$5) and C $$(-1$$ ,$$0$$,$$2)$$ be three points, then equation of a plane parallel to the plane ABC which is at distance  $$2$$ is

Infinity Learn · Accepted Answer

$$A(1$$,$$1$$,$$1)$$,$$B(2$$,$$3$$,$$5)$$,$$C($$−$$1$$,$$0$$,$$2)$$ direction ratios of AB are <$$1$$,$$2$$,$$4$$⟩ Direction ratios of AC are <−$$2$$,−$$1$$,$$1$$>. Therefore, direction ratios of normal to plane ABC are <$$2$$,−$$3$$, $$1$$> As a result, equation of the plane ABC is $$2x$$−$$3y+z=0$$. Let the equation of the required plane be $$2x$$−$$3y+z=k$$. Then $$k4+9+1=2$$ or $$k=$$±$$214$$ Hence, equation of the required plane is $$2x$$−$$3y+z+214=0$$

Let A (1, 1, 1), B (2,3,5) and C (-1 ,0,2) be three points, then equation of a plane parallel to the plane ABC which is at distance 2 is

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