If the vertices of a triangle are $$1$$,$$2$$,$$3$$,$$3$$,$$4$$,−$$1$$,$$5$$,−$$4$$,$$2and$$ the area of triangle is λ , then the number of positive integral divisors of λ$$26+$$λ$$2+253$$ is

Question

If the vertices of a triangle are $$1$$,$$2$$,$$3$$,$$3$$,$$4$$,−$$1$$,$$5$$,−$$4$$,$$2and$$ the area of triangle is λ , then the number of positive integral divisors of  λ$$26+$$λ$$2+253$$ is

Infinity Learn · Accepted Answer

The area of the triangle formed by the vertices A, B, C is $$12AB$$¯×AC¯Here,             AB¯$$=OB$$¯−OA¯$$=3i+4j$$−k−i−$$2j$$−$$3k=2i+2j$$−$$4kAC=OC$$−$$OA=5i$$−$$4j+2k$$−i−$$2j$$−$$3k=4i$$−$$6j$$−kHence, the area of the triangle is $$12AB$$¯×AC¯$$=12ijk22$$−$$44$$−$$6$$−$$1=12i$$−$$2$$−$$24$$−j−$$2+16+k$$−$$12$$−$$8=12$$−$$26i$$−$$14j$$−$$20k=318$$  Consider ,        λ$$26+$$λ$$2+253=3186+318+253=53+3433=60=22$$⋅$$3$$⋅$$5$$    The number of integral divisors of the above number is  $$2+1$$⋅$$1+1$$⋅$$1+1=12$$

If the vertices of a triangle are $(1, 2, 3), (3, 4, - 1), (5, - 4, 2)$ and the area of triangle is $λ$ , then the number of positive integral divisors of $(\frac{λ^{2}}{6} + \sqrt[3]{λ^{2} + 25})$ is

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If the vertices of a triangle are 1,2,3,3,4,−1,5,−4,2and the area of triangle is λ , then the number of positive integral divisors of λ26+λ2+253 is

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If the vertices of a triangle are $(1, 2, 3), (3, 4, - 1), (5, - 4, 2)$ and the area of triangle is $λ$ , then the number of positive integral divisors of $(\frac{λ^{2}}{6} + \sqrt[3]{λ^{2} + 25})$ is