and $x - y = 0$ .…$(ii)$
After solving $(i)$ and $(ii),$ $x = 2,\,y = 2$
$\therefore$ $2x + z = 7$ $⇒$ $z = 3$ and $2z + w = 10$ $⇒$ $w = 4$.
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$\left[\begin{array}{lll}a & b & c\end{array}\right]\left[\begin{array}{lll}1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$................$(E)$
$1.$ If the point $P(a, b, c)$, with reference to $( E )$, lies on the plane $2 x+y+z=1$, then the value of $7 a+b+c$ is
$(A)$ $0$ $(B)$ $12$ $(C)$ $7$ $(D)$ $6$
$2.$ Let $\omega$ be a solution of $x^3-1=0$ with $\operatorname{Im}(\omega)>0$. If $a=2$ with $b$ and $c$ satisfying $( E )$, then the value of $\frac{3}{\omega^a}+\frac{1}{\omega^b}+\frac{3}{\omega^c}$ is equal to
$(A)$ $-2$ $(B)$ $2$ $(C)$ $3$ $(D)$ $-3$
$3.$ Let $b=6$, with $a$ and $c$ satisfying (E). If $\alpha$ and $\beta$ are the roots of the quadratic equation $a x^2+b x+c=0$, then
$\sum_{n=0}^{\infty}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)^n$ is
$(A)$ $6$ $(B)$ $7$ $(C)$ $\frac{6}{7}$ $(D)$ $\infty$
Give the answer question $1,2$ and $3.$
$\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5} \text { and } \frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$ is :