The equation of the plane through the points p(0, 1, 1), q(1, 0, 1), and
r(1, 1, 0) is x + y + z = 2 .
According to the question
Let the points
p(0, 1, 1) = (x1,y1,z1)
q(1, 0, 1)= (x2,y2,z2)
r(1, 1, 0)= (x3,y3,z3)
now,
The equation of the plane through the points
i.e
The formula of equation of a plane passing through three non collinear points
[tex]\left[\begin{array}{ccc}x-x1&y-y1&z-z1\\x2-x1&y2-y1&z2-z1\\x3-x1&y3-y1&z3-z1\end{array}\right] = 0[/tex]
by substituting the values in the formula of equation of a plane
[tex]\left[\begin{array}{ccc}x-0&y-1&z-1\\1-0&0-1&1-1\\1-0&1-1&0-1\end{array}\right] = 0[/tex]
[tex]\left[\begin{array}{ccc}x&y-1&z-1\\1&-1&0\\1&0&-1\end{array}\right] = 0[/tex]
x(1) - (y-1)(-1) + (z-1)(1) = 0
x + y - 1 + z - 1 = 0
x + y + z = 2
Hence, the equation of the plane through the points p(0, 1, 1), q(1, 0, 1), and r(1, 1, 0) is x + y + z = 2 .
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