Consider three planes $$2x+py+6z=8$$, x $$+$$ $$2y$$ $$+$$ qz $$=$$ $$5$$ and x $$+$$ y $$+$$ $$3z$$ $$=$$ $$4Three$$ planes intersect at a point ifthree planes do not have any common point of intersection if

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The given system of equations is $$2x+py+6z=8x+2y+qz=5x+y+3z=4$$Δ$$=2p612q113=(2$$−$$p)(3$$−$$q)$$ By Cramer's rule, if Δ≠$$0$$, i.e., p≠$$2$$ and q≠$$3$$, the  system has a unique solution.  If $$p=2$$ or $$q=3$$,Δ$$=0$$, then if Δ$$x=$$Δ$$v=$$Δ$$z=0$$, the system has infinite solutions and if any one of Δx,Δy and Δz≠$$0$$  the system has no solutionnow,Δ$$x=8p652q413=30$$−$$8q$$−$$15p+4pq=(4q$$−$$15)$$⋅(p$$−$$2)=28615q143=$$−$$8q+8q=0$$Δ$$y=2p8125114=p$$−$$2Thus$$, if $$p=2$$,Δ$$x=$$Δ$$y=$$Δ$$z=0$$  for all q∈R  the system has infinite solutions. if p≠$$2$$,$$q=3$$ and Δz≠$$0$$  then the system has no solution. Hence the system has $$(i) no solution if p≠$$2$$  and q $$=$$ $$3$$,(ii) a unique solution if p≠$$2$$ and q≠$$2$$ and (iii) infinite solutions if p $$=$$  $$2$$ and q∈RThe given system of equations is $$2x+py+6z=8x+2y+qz=5x+y+3z=4$$Δ$$=2p612q113=(2$$−$$p)(3$$−$$q)$$ By Cramer's rule, if Δ≠$$0$$, i.e., p≠$$2$$ and q≠$$3$$, the  system has a unique solution.  If $$p=2$$ or $$q=3$$,Δ$$=0$$, then if Δ$$x=$$Δ$$v=$$Δ$$z=0$$, the system has infinite solutions and if any one of Δx,Δy and Δz≠$$0$$  the system has no solutionnow,Δ$$x=8p652q413=30$$−$$8q$$−$$15p+4pq=(4q$$−$$15)$$⋅(p$$−$$2)=28615q143=$$−$$8q+8q=0$$Δ$$y=2p8125114=p$$−$$2Thus$$, if $$p=2$$,Δ$$x=$$Δ$$y=$$Δ$$z=0$$  for all q∈R  the system has infinite solutions. if p≠$$2$$,$$q=3$$ and Δz≠$$0$$  then the system has no solution. Hence the system has $$(i) no solution if p≠$$2$$  and q $$=$$ $$3$$,(ii) a unique solution if p≠$$2$$ and q≠$$2$$ and (iii) infinite solutions if p $$=$$  $$2$$ and q∈RThe given system of equations is $$2x+py+6z=8x+2y+qz=5x+y+3z=4$$Δ$$=2p612q113=(2$$−$$p)(3$$−$$q)$$ By Cramer's rule, if Δ≠$$0$$, i.e., p≠$$2$$ and q≠$$3$$, the  system has a unique solution.  If $$p=2$$ or $$q=3$$,Δ$$=0$$, then if Δ$$x=$$Δ$$v=$$Δ$$z=0$$, the system has infinite solutions and if any one of Δx,Δy and Δz≠$$0$$  the system has no solutionnow,Δ$$x=8p652q413=30$$−$$8q$$−$$15p+4pq=(4q$$−$$15)$$⋅(p$$−$$2)=28615q143=$$−$$8q+8q=0$$Δ$$y=2p8125114=p$$−$$2Thus$$, if $$p=2$$,Δ$$x=$$Δ$$y=$$Δ$$z=0$$  for all q∈R  the system has infinite solutions. if p≠$$2$$,$$q=3$$ and Δz≠$$0$$  then the system has no solution. Hence the system has $$(i) no solution if p≠$$2$$  and q $$=$$ $$3$$,(ii) a unique solution if p≠$$2$$ and q≠$$2$$ and (iii) infinite solutions if p $$=$$  $$2$$ and q∈R

Consider three planes 2x+py+6z=8, x + 2y + qz = 5 and x + y + 3z = 4Three planes intersect at a point ifthree planes do not have any common point of intersection if

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