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

$$D=|\begin{matrix}1 & 2 & -3 \\ 2 & 6 & -11 \\ 1 & -2 & 7\end{matrix}|$$D=∣121​26−2​−3−117​∣

$$=20-2(25)-3(-10)$$=20−2(25)−3(−10)

$$=20-50+30=0$$=20−50+30=0

$${D}_{1}=|\begin{matrix}a & 2 & -3 \\ b & 6 & -11 \\ c & -2 & 7\end{matrix}|$$D1​=∣abc​26−2​−3−117​∣

$$=20a-2(7b+11c)-3(-2b-6c)$$=20a−2(7b+11c)−3(−2b−6c)

$$=20a-14b-22c+6b+18c$$=20a−14b−22c+6b+18c

$$=20a-8b-4c$$=20a−8b−4c

$$=4(5a-2b-c)$$=4(5a−2b−c)

$${D}_{2}=|\begin{matrix}1 & a & -3 \\ 2 & b & -11 \\ 1 & c & 7\end{matrix}|$$D2​=∣121​abc​−3−117​∣

$$=7b+11c-a(25)-3(2c-b)$$=7b+11c−a(25)−3(2c−b)

$$=7b+11c-25a-6c+3b$$=7b+11c−25a−6c+3b

$$=-25a+10b+5c$$=−25a+10b+5c

$$=-5(5a-2b-c)$$=−5(5a−2b−c)

$${D}_{3}=|\begin{matrix}1 & 2 & a \\ 2 & 6 & b \\ 1 & -2 & c\end{matrix}|$$D3​=∣121​26−2​abc​∣

$$=6c+2b-2(2c-b)-10a$$=6c+2b−2(2c−b)−10a

$$=-10a+4b+2c$$=−10a+4b+2c

$$=-2(5a-2b-c)$$=−2(5a−2b−c)

for infinite solution

$$D={D}_{1}={D}_{2}={D}_{3}=0$$D=D1​=D2​=D3​=0

$$\Rightarrow 5a=2b+c$$⇒5a=2b+c