If x+3z+42y−74x+6a−10b−33bz+2c=063y−22x−32c+22b+4−210then, the values of a,b,c,x,y and z respectively are

Since x+3z+42y−74x+6a−10b−33bz+2c=063y−22x−32c+22b+4−210∴x+3=0⇒x=−3b−3=2b+4⇒b=−7z+4=6⇒z=2a−1=−3⇒a=−22y−7=3y−2⇒y=−52c+2=0⇒c=−1∴x=−3,y=−5,z=2,a=−2,b=−7,c=−1

If x+3z+42y−74x+6a−10b−33bz+2c=063y−22x−32c+22b+4−210then, the values of a,b,c,x,y and z respectively are

Since x+3z+42y−74x+6a−10b−33bz+2c=063y−22x−32c+22b+4−210∴x+3=0⇒x=−3b−3=2b+4⇒b=−7z+4=6⇒z=2a−1=−3⇒a=−22y−7=3y−2⇒y=−52c+2=0⇒c=−1∴x=−3,y=−5,z=2,a=−2,b=−7,c=−1

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If $[\begin{array}{ccc} x + 3 & z + 4 & 2 y - 7 \\ 4 x + 6 & a - 1 & 0 \\ b - 3 & 3 b & z + 2 c \end{array}] = [\begin{array}{ccc} 0 & 6 & 3 y - 2 \\ 2 x & - 3 & 2 c + 2 \\ 2 b + 4 & - 21 & 0 \end{array}]$
then, the values of $a, b, c, x, y$ and $z$ respectively are

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– 2, – 7, – 1, – 3, – 5, – 2

2, 7, 1, 3, 5, – 2

1, 3, 4, 2, 8, 9

– 1, 3, –2, –7, 4, 5

answer is A.

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Detailed Solution

Since $[\begin{array}{ccc} x + 3 & z + 4 & 2 y - 7 \\ 4 x + 6 & a - 1 & 0 \\ b - 3 & 3 b & z + 2 c \end{array}]$

$= [\begin{array}{ccc} 0 & 6 & 3 y - 2 \\ 2 x & - 3 & 2 c + 2 \\ 2 b + 4 & - 21 & 0 \end{array}]$

$\begin{array}{l} ∴ x + 3 = 0 \Rightarrow x = - 3 \\ b - 3 = 2 b + 4 \Rightarrow b = - 7 \\ z + 4 = 6 \Rightarrow z = 2 \\ a - 1 = - 3 \Rightarrow a = - 2 \\ 2 y - 7 = 3 y - 2 \Rightarrow y = - 5 \\ 2 c + 2 = 0 \Rightarrow c = - 1 \\ ∴ x = - 3, y = - 5, z = 2, a = - 2, b = - 7, c = - 1 \end{array}$