Question

Let P be the image of the point (3, 1, 7) with respect to the plane x - y + z = 3. Then, the equation of the plane passing through P and containing the straight line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ is

• A. x + y - 3z = 0
• B. 3x + z = 0
• C. x - 4y + 7z = 0
• D. 2x - y = 0

Answer 1

The correct option is C

x - 4y + 7z = 0

Let image of $Q(3,1,7)$ w.r.t. x - y + z = 3 be $P(\alpha ,\beta ,\gamma ).\therefore \frac{\alpha -3}{1}=\frac{\beta -1}{-1}=\frac{\gamma -7}{1}=\frac{-2(3-1+7-)}{1^2+(-1{)}^2+(1{)}^2}\Rightarrow \alpha -3=1-\beta =\gamma -7=-4\therefore \alpha =-1,\beta =5,\gamma =3$

Hence, the image of Q(3, 1, 7) is P(-1, 5, 3).
To find equation of plane passing through P(-1, 5, 3) and containing $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$

$\Rightarrow \left|\begin{array}{ccc}x-0y-0z-0& & \\ 1-02-01-0& & \\ -1-05-03-0& & \end{array}\right|=0\Rightarrow x(6-5)-y(3+1)+z(5+2)=0x-4y+7z=0$