Question

Let \(P\) be the plane such that it contains the straight line \[ \frac{x-1}{2}=\frac{y-3}{3}=\frac{z+2}{1} \] and is perpendicular to the plane \[ x+2y+3z=4 \] Let \(P_1\) be the plane which passes through the point \((4,2,2)\) and is parallel to \(P\). Then which of the following statements is (are) TRUE?

• A. The equation of the plane \(P\) is 7x-5y+z=-10
• B. The distance between the planes \(P\) and \(P_1\) is \(30\)
• C. The distance of the plane \(P\) from the origin is \(2\sqrt3\)
• D. The acute angle between the plane \(P\) and the plane 2x+2y+z=3 is \[\cos^{-1}\left(\frac1{3\sqrt3}\right)\]

Hint:

If two planes are perpendicular, then: \[ \vec n_1\cdot\vec n_2=0 \] Distance between parallel planes: \[ \frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}} \]

Answer 1

The Correct Option is A

Step 1: Find the normal vector of plane \(P\). 
Direction ratios of the given line: \[ (2,3,1) \] Normal vector of plane \[ x+2y+3z=4 \] is: \[ (1,2,3) \] Since plane \(P\) contains the line and is perpendicular to the given plane, its normal vector is perpendicular to both \[ (2,3,1) \] and \[ (1,2,3). \] Thus, \[ \vec n= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 1 \\ 1 & 2 & 3 \end{vmatrix} \] \[ \vec n=(7,-5,1) \] 

Step 2: Find equation of plane \(P\). 
Plane passes through point: \[ (1,3,-2) \] Using: \[ 7(x-1)-5(y-3)+(z+2)=0 \] \[ 7x-7-5y+15+z+2=0 \] \[ 7x-5y+z+10=0 \] \[ 7x-5y+z=-10 \] Therefore: \[ \Rightarrow \mathrm{Option\ (A)\ is\ Correct} \] 

Step 3: Check Option (B). 
Plane \(P_1\) parallel to \(P\) through: \[ (4,2,2) \] Equation: \[ 7(x-4)-5(y-2)+(z-2)=0 \] \[ 7x-5y+z=20 \] Distance between planes: \[ \frac{|20-(-10)|}{\sqrt{7^2+(-5)^2+1^2}} \] \[ =\frac{30}{\sqrt{75}} \] \[ =\frac{30}{5\sqrt{3}} \] \[ =2\sqrt{3} \] Thus: \[ \Rightarrow \mathrm{Option\ (B)\ is\ Incorrect} \] 

Step 4: Check Option (C). 
Distance of plane \[ 7x-5y+z+10=0 \] from origin: \[ \frac{|10|}{\sqrt{75}} \] \[ =\frac{2}{\sqrt{3}} \] not \[ 2\sqrt{3} \] Therefore: \[ \Rightarrow \mathrm{Option\ (C)\ is\ Incorrect} \] 

Step 5: Check Option (D). 
Normal vector of \[ 2x+2y+z=3 \] is: \[ (2,2,1) \] Angle between planes equals angle between normals. Thus: \[ \cos\theta= \frac{|(7)(2)+(-5)(2)+(1)(1)|} {\sqrt{75}\sqrt{9}} \] \[ = \frac{|14-10+1|} {5\sqrt{3}\cdot3} \] \[ = \frac{5}{15\sqrt{3}} \] \[ = \frac{1}{3\sqrt{3}} \] Hence: \[ \theta= \cos^{-1}\left(\frac{1}{3\sqrt{3}}\right) \] Therefore: \[ \Rightarrow \mathrm{Option\ (D)\ is\ Correct} \] 

Step 6: Identify the correct options. 
Hence: \[ \boxed{\mathrm{(A)\ and\ (D)}} \]