Consider the following system of equations: x | vedclass

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(B) Let the given equations be:$P_{1}: x+2y-3z=a$$P_{2}: 2x+6y-11z=b$$P_{3}: x-2y+7z=c$First,we calculate the determinant of the coefficient matrix $D

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has infinite number of solutions when $5a=2b+c$

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(B) Let the given equations be:$P_{1}: x+2y-3z=a$$P_{2}: 2x+6y-11z=b$$P_{3}: x-2y+7z=c$First,we calculate the determinant of the coefficient matrix $D$:$D = \begin{vmatrix} 1 & 2 & -3 \ 2 & 6 & -11 \ 1 & -2 & 7 \end{vmatrix} = 1(42-22) - 2(14+11) - 3(-4-6) = 20 - 50 + 30 = 0$.Since $D=0$,the system does not have a unique solution.Now,observe the linear combination of the equations:$2P_{1} + P_{3} = 2(x+2y-3z) + (x-2y+7z) = 2x+4y-6z + x-2y+7z = 3x+2y+z$ (This does not match $P_{2}$ directly).Let us check $5P_{1} = 2P_{2} + P_{3}$:$2P_{2} + P_{3} = 2(2x+6y-11z) + (x-2y+7z) = 4x+12y-22z + x-2y+7z = 5x+10y-15z = 5(x+2y-3z) = 5P_{1}$.Thus,if $5a = 2b + c$,the third equation is a linear combination of the first two,meaning the planes are consistent and share a common line of intersection.Therefore,the system has an infinite number of solutions when $5a = 2b + c$.

Consider the following system of equations: $x+2y-3z=a$,$2x+6y-11z=b$,and $x-2y+7z=c$,where $a, b$,and $c$ are real constants. Then the system of equations:

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