Question

Find the equation of the plane through the intersection of the plane \[3x - y + 2z - 4 = 0,x + y + z - 2 = 0\] and the point \[(2,2,1)\] .
A.\[7x + 5y + 4z + 8 = 0\]
B. \[7x + 5y + 4z - 8 = 0\]
C. \[7x - 5y + 4z - 8 = 0\]
D. None of these

Answer 1

Hint: First write the equation of the plane passing though intersection of two planes as \[(3x - y + 2z - 4) + a(x + y + z - 2) = 0\], then substitute 2 for x, 2 for y ad 1 for z in the equation \[(3x - y + 2z - 4) + a(x + y + z - 2) = 0\] and obtain the value of a.

Formula used:
The equation of the plane passing through the intersection of two planes \[ax + by + cz - d = 0,px + qy + rz - s = 0\] is \[(ax + by + cz - d) + m(px + qy + rz - s) = 0\], where m is any real number.

Complete step by step solution:
The given equations of the planes are \[3x - y + 2z - 4 = 0,x + y + z - 2 = 0\].
Therefore, the equation of the plane passing through the intersection of these two planes is,
 \[(3x - y + 2z - 4) + a(x + y + z - 2) = 0 - - - - (1)\]
Now, it is given that the plane also passes through the point \[(2,2,1)\].
So,
\[(3.2 - 2 + 2.1 - 4) + a(2 + 2 + 1 - 2) = 0\]
\[(8 - 6) + a.3 = 0\]
\[3a = - 2\]
\[a = - \dfrac{2}{3}\]
Substitute \[a = - \dfrac{2}{3}\] in equation (1) to obtain the required solution.
\[(3x - y + 2z - 4) - \dfrac{2}{3}(x + y + z - 2) = 0\]
\[3(3x - y + 2z - 4) - 2(x + y + z - 2) = 0\]
\[9x - 3y + 6z - 12 - 2x - 2y - 2z + 4 = 0\]
\[7x - 5y + 4z - 8 = 0 \]

Therefore, the correct option is C.

Note: Students can solve the equation (1) and form the equation as \[(3 + a)x + (a - 1)y + (2 + a)z - (4 + 2a) = 0\] then substitute 2 for x, 2 for y ad 1 for z to obtain the value of a as \[ - \dfrac{2}{3}\] . Then substitute \[a = - \dfrac{2}{3}\] in the equation \[(3 + a)x + (a - 1)y + (2 + a)z - (4 + 2a) = 0\] and solve to obtain the required solution.