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If |{:(x1,,y1,,1),(x2,,y2,,1),(x3,,y3,,1...

If `|{:(x_1,,y_1,,1),(x_2,,y_2,,1),(x_3,,y_3,,1):}|=|{:(a_1,,b_1,,1),(a_2,,b_2,,1),(a_3,,b_3,,1):}|`
then the two triangles with vertices `(x_1,y_1),(x_2,y_2),(x_3,y_3)` and `(a_1,b_1), (a_2,b_2), (a_3,b_3)` are

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Consider the following system of equations a_1x+b_1y+c_1z=d_1, a_2x+b_2y+c_2z=d_2, a_3x+b_3y+c_3z=d_3 Let /_\= |(a_1,b_1,c_1), (a_2,b_2,c_2), (a_3,b_3,c_3)|, /_\_1= |(d_1,b_1,c_1), (d_2,b_2,c_2), (d_3,b_3,c_3)|, ,/_\_2=|(a_1,d_1,c_1), (a_2,d_2,c_2), (a_3,d_3,c_3)|,, /_\_3=|(a_1,b_1,cd_1), (a_2,b_2,d_2), (a_3,b_3,d_3)| , The given system of equations will have i. unique solution if /_\!=0 ii. infinitely many solutions if /_\=/_\_1=/_\_3=0 . iii. no solution if /_\=0 and any of /_\_1, /_\_2, /_\_3 is none zero. On the basis of above informatioin answer thefollowing questions for the following system of linear equations. . 2x+ay+6z=8, x+2y+bz=5, x+y+3z=4 The given system of equatioin has unique solution if (A) a=2,b=2 (B) a!=2,b=3 (C) a!=2, b!=3 (D) a=2,b!=3

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