Question

Find the point where the line of intersection of the planes $x-2y+z=1$ and $x+2y-2z=5$, intersects the plane $2x+2y+z+6=0$

• A. $(1,-2,-4)$
• B. $(0,0,-6)$
• C. $(1,0,-8)$
• D. $(-1,-1,-2)$

Answer 1

Answer: (A)

$(1,-2,-4)$

Given planes are $x-2y+z=1...\left(i\right)$
$x+2y-2z=5...\left(ii\right)$
and $2x+2y+z=-6...\left(iii\right)$
Add $\left(i\right)+\left(ii\right)+\left(iii\right)$
$4x+2y=0\Rightarrow y=-2x...\left(iv\right)$
From equations $\left(iii\right)-\left(i\right)$
$x+4y=-7...\left(v\right)$
from (iv) and (v) we get
$x=1,y=-2$
Put in(i) we get $z=-4$
So point of intersection is $(1,-2,-4)$