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Let $a, b$ and $c$ be three real numbers satisfying
$\left[\begin{array}{lll}a & b & c\end{array}\right]\left[\begin{array}{lll}1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$................$(E)$
$1.$ 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 $3.$
$(D,A,B)$
$(A,B,C)$
$(D,D,B)$
$(C,B,B)$
Solution
$1.$ Given
$a+8 b+7 c=0$
$9 a+2 b+3 c=0$
$\text { and } a+b+c=0$
Solving, we get $b=6 a, c=-7 a$
Since the point $P$ lies on the plane, so
$2 a+b+c=1$
$2 a+6 a-7 a=1$
$a=1$
Thus, $b =6$ and $c =-7$
Now the value of $7 a+b+c$
$=7+6-7$
$=6$
$2.$ The correct option is $A -2$
On multiplying matrices, we have
$\left[\begin{array}{lll}a+8 b+7 c & 9 a+2 b+3 c & 7 a+7 b+7 c\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$
On solving the 3 equations, we get $b=6 a$ and $c=-7 a$
$\text { If } a=2 \text {, we get, } b=12, c=-14$
$\therefore \frac{3}{\omega^a}+\frac{1}{\omega^b}+\frac{3}{\omega^c}$
$\Rightarrow \frac{3}{\omega^2}+\frac{1}{\omega^{12}}+\frac{3}{\omega^{-14}}=\frac{3}{\omega^2}+1+3 \omega^2$
$=3 \omega+1+3 \omega^2$
$=1+3\left(\omega+\omega^2\right)=1-3=-2$
$3.$ Since, $a , b$ and c satisfies
$\left[\begin{array}{lll} a & b & c \end{array}\right]\left[\begin{array}{lll}1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$
So, we get the equations
$a+8 b+7 c=0$
$9 a+2 b+3 c=0$
$7 a+7 b+7 c=0 \text { or } a+b+c=0$
Since, $b =6$, so the equations become
$a+48+7 c=0 \ldots$
$9 a+12+3 c=0 .$
$a+6+c=0 \ldots . .3$
Subtracting $(3)$ from $(1)$, we get
$c=-7$
Using equation $(3)$, we get $a =1$.
So, we have $a =1, b=6, c =-7$
