Question 1008762
Question 1008730
<pre>
z = ax + by + c

Substitute (x, y, z) = (-1, 0, -4)

        -4 = a(-1) + b(0) + c

        -4 = -a + c

     a - c = 4

Substitute (x, y, z) = (2, 1, 8)

          8 = a(2) + b(1) + c

          8 = 2a + b + c

-2a - b - c = -8

 2a + b + c = 8

Substitute (x, y, z) = (-1, 1, -1)

        -1 = a(-1) + b(1) + c

        -1 = -a + b + c

 a - b - c = 1


So we have this system of equations:

{{{system(matrix(3,7,

    a,  ""  , "" ,""-"", c, ""="", 4,
 2a,""+"", b, ""+"", c,""="", 8,
 a,""-"", b, ""-"", c, ""="", 1))}}}

That's easy to solve for a, b, and c.

The substitute the vales you get for a,b, and c

in z = ax + by + c to get your answer.

Want to see your plane graphed in 3D?  Look below.

It's the plane in which the green triangle is in. 


{{{drawing(400,400,-10,10,-10,10,

red(line(-7,2,7,-2), line(-7,-4,7,4), line(0,-8,0,8)),


line(7/sqrt(65),4/sqrt(65),7/sqrt(65),4/sqrt(65)-4),

line(-14/sqrt(65),-8/sqrt(65),-14/sqrt(65)+7/sqrt(53),-8/sqrt(65)-2/sqrt(53)),

line(-14/sqrt(65)+7/sqrt(53),-8/sqrt(65)-2/sqrt(53),



-14/sqrt(65)+7/sqrt(53),-8/sqrt(65)-2/sqrt(53)+8),

line(7/sqrt(65),4/sqrt(65),7/sqrt(65)+7/sqrt(53),4/sqrt(65)-2/sqrt(53)),


line(7/sqrt(65)+7/sqrt(53),4/sqrt(65)-2/sqrt(53),7/sqrt(65)+7/sqrt(53),4/sqrt(65)-2/sqrt(53)-1),
red(locate(-7,-4,x),locate(7,4,-x),locate(-8,2.4,-y), locate(7,-2,y),



locate(0,8.8,z),locate(0,-8,-z)),



line(0,0,7/sqrt(65),4/sqrt(65)),

locate(.8,-3.6,"(-1, 0, -4)"),

locate(1.8,-.8,"(-1, 1, -1)"),

locate(-.81,6.82,"(2, 1, 8)"),

green(triangle(-14/sqrt(65)+7/sqrt(53),-8/sqrt(65)-2/sqrt(53)+8,

7/sqrt(65)+7/sqrt(53),4/sqrt(65)-2/sqrt(53)-1,

7/sqrt(65),4/sqrt(65)-4))


 )}}}


Edwin</pre>