Question

What is the equation of the plane passing through the point \[\left( {1,1,1} \right)\] and the intersection of the planes \[x + y + z = 6\] and \[2x + 3y + 4z + 5 = 0\]?
A. \[20x + 23y + 26z - 69 = 0\]
B. \[20x + 23y + 26z + 69 = 0\]
C. \[23x + 20y + 26z - 69 = 0\]
D. None of these

Answer 1

Hint: First, find the equation of the required plane by using the formula of the equation of the plane passing through the intersection of the two planes. Then, substitute the coordinates of the given point in that equation and solve it to find the value of the variable. In the end, substitute the value of the variable in the first equation and solve the equation to get the required answer.

Formula used: The equation of the plane passing through the intersection of the two planes \[a{x_1} + b{y_1} + c{z_1} + {d_1} = 0\] and \[a{x_2} + b{y_2} + c{z_2} + {d_2} = 0\] is: \[\left( {a{x_1} + b{y_1} + c{z_1} + {d_1}} \right) + \lambda \left( {a{x_2} + b{y_2} + c{z_2} + {d_2}} \right) = 0\]

Complete step by step solution: Given: The plane passing through the point \[\left( {1,1,1} \right)\] also passes through the intersection of planes \[x + y + z = 6\] and \[2x + 3y + 4z + 5 = 0\].

Let’s find out the equation of the plane.
Apply the formula for the equation of the plane passing through the intersection of the two planes.
So, the equation of the required plane:
\[\left( {x + y + z - 6} \right) + \lambda \left( {2x + 3y + 4z + 5} \right) = 0\] \[.....\left( 1 \right)\]
It is given that, the plane passes through the point \[\left( {1,1,1} \right)\].
So, the point satisfies the equation of the plane.
We get,
\[\left( {1 + 1 + 1 - 6} \right) + \lambda \left( {2\left( 1 \right) + 3\left( 1 \right) + 4\left( 1 \right) + 5} \right) = 0\]
\[ \Rightarrow - 3 + \lambda \left( {2 + 3 + 4 + 5} \right) = 0\]
\[ \Rightarrow - 3 + 14\lambda = 0\]
\[ \Rightarrow 14\lambda = 3\]
\[ \Rightarrow \lambda = \dfrac{3}{{14}}\] \[.....\left( 2 \right)\]

Now substitute the equation \[\left( 2 \right)\] in the equation \[\left( 1 \right)\].
 \[\left( {x + y + z - 6} \right) + \dfrac{3}{{14}}\left( {2x + 3y + 4z + 5} \right) = 0\]
\[ \Rightarrow 14\left( {x + y + z - 6} \right) + 3\left( {2x + 3y + 4z + 5} \right) = 0\]
\[ \Rightarrow 14x + 14y + 14z - 84 + 6x + 9y + 12z + 15 = 0\]
\[ \Rightarrow 20x + 23y + 26z - 69 = 0\]
Thus, the equation of the plane passing through the point \[\left( {1,1,1} \right)\] and the intersection of the planes \[x + y + z = 6\] and \[2x + 3y + 4z + 5 = 0\] is \[20x + 23y + 26z - 69 = 0\].

Thus, Option (A) is correct.

Note: Students often get confused while solving the equation \[\left( {x + y + z - 6} \right) + \dfrac{3}{{14}}\left( {2x + 3y + 4z + 5} \right) = 0\]. They simplify the equation as \[\left( {x + y + z - 6} \right) + 3\left( {2x + 3y + 4z + 5} \right) = 0 \times 14\], which is an incorrect method. So, to simplify the equation multiply both sides by \[14\]. i.e., \[14\left( {x + y + z - 6} \right) + 3\left( {2x + 3y + 4z + 5} \right) = 0 \times 14\].