Correct option is (4) (2, –4, 1)
Explaination::
From question, the equations of planes given are:
2x – y + 2z – 4 = 0
X + 2y + 2z – 2 = 0
The equation of the planes bisecting the angles between two given planes a_{1}x + b_{1}y + c_{1}z + d_{1 }= 0 and a_{2}x + b_{2}y + c_{2}z + d_{2 }= 0 is given as:
\(\frac{{{a_1}x + {b_1}y + {c_1}z + {d_1}}}{{\sqrt {\left( {a_1^2 + b_1^2 + c_1^2} \right)} }} = \pm \frac{{{a_2}x + {b_2}y + {c_2}z + {d_2}}}{{\sqrt {\left( {a_2^2 + b_2^2 + c_2^2} \right)} }}\)
Now, for this problem, the equation of the planes bisecting the angles is:
\(\Rightarrow \frac{{2x - y + 2z - 4}}{{\sqrt {{2^2} + {{( - 1)}^2} + {2^2}} }} = \pm \left( {\frac{{x + 2y + 2z - 2}}{{\sqrt {{1^2} + {2^2} + {2^2}} }}} \right)\)
\(\Rightarrow \frac{{2x - y + 2z - 4}}{{\sqrt 9 }} = \pm \left( {\frac{{x + 2y + 2z - 2}}{{\sqrt 9 }}} \right)\)
\(\Rightarrow \frac{{2x - y + 2z - 4}}{3} = \pm \left( {\frac{{x + 2y + 2z - 2}}{3}} \right)\)
⇒ 2x – y + 2z – 4 = ±(x + 2y + 2z – 2)
Case I: take positive sign
⇒ 2x – y + 2z – 4 = x + 2y + 2z - 2
⇒ x – 3y – 2 = 0 ----(1)
Case II: take negative sign
⇒ 2x – y + 2z – 4 = -(x + 2y + 2z - 2)
⇒ 2x – y + 2z – 4 = -x – 2y – 2z + 2
⇒ 3x + y + 4z – 6 = 0 ----(2)
Now, we need to find which point in the given options satisfies the obtained equation.
The point (2, -4, 1) satisfies equation (2)