Question

If $x=a,y=b,z=c$ is a solution of the system of linear equations $x+8y+7z=0,9x+2y+3z=0,x+y+z=0$ such that the point$(a,b,c)$ lies on the plane $x+2y+z=6$, then $2a+b+c$ equals.

• A. $0$
• B. $2$
• C. $1$
• D. $-1$

Answer 1

The correct option is C $1$

$x+8y+7z=0$ ........(1)
$9x+2y+3z=0$ .........(2)
$x+y+z=0$ ........(3)
Multiply eqn. (3) by 3 & subtract from eqn. (2)
$9x+2y+3z=0$
$3x\pm 3y\pm 3z=0$
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$6x-y=0$
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$y=6x$ .........(4)
Substitute in Eqn (1). we get
$x+8\left(6x\right)+7z=0$
$49x+7z=0$
$7x+z=0$
$z=-7x$ .........(5)
$\Rightarrow x=a$, $y=b.z=c$
is the solutions of the system of linear equation
$\therefore b=6a$ and $c=-7a$ .......(6)
Also, (a, b, c) lies on the plane $x+2y+z=6$
$\therefore a+2b+c=6$ ........(7)
Substitute eq. (6) in eq(7)
$\therefore a+2\left(6a\right)-7a=6$
$\therefore a+12a-7a=6$
$\therefore 6a=6$
$\therefore a=1$
So, $b=6$ and $c=-7$
$\therefore 2a+b+c=2\left(1\right)+6-7=1$.