The plane passing through the points (1,2,1), (2,1,2) and parallel to the line, 2x=3y,z=1, also passes through the point:
Solutions · 1
Official solution
Equation of plane passing through (2, 1, 2)
a(x - 2) + b(y - 1) + c(z - 2) = 0 ......(1)
As point (1, 2, 1) also passes through the plane, so it satisfy the equation,
a(1 - 2) + b(2 - 1) + c(1 - 2) = 0
\[ \Rightarrow \] -a + b - c = 0 ....(2)
Given line 2x = 3y and z = 1,
So, symmetric form of the line
\[{x \over 3} = {y \over 2} = {{z - 1} \over 0}\]
\[ \therefore \] Direction ratio of this line is (3, 2, 0) and Direction ration of plane = (a, b, c)
As plane is parallel to the line so the normal of the plane is perpendicular to the line.
\[ \therefore \] Dot product of direction ratio = 0
3a + 2b + 0(c) = 0 .....(3)
Equation of plane,
\[\left| {x−2−13y−112z−2−10 } \right| = 0\]
\[ \Rightarrow 3(1 - y + 2 - z) - 2( - x + 2 + z - 2) = 0\]
\[ \Rightarrow 9 - 3y - 3z + 2x - 2z = 0\]
\[ \Rightarrow 2x - 3y - 5z + 9 = 0\]
By checking all options you can see (-2, 0, 1) satisfy the equation.