Question

Let $P=0$ be the equation of a plane passing through the line of intersection of the planes $2x-y=0$ and $3z-y=0$ and perpendicular to the plane $4x+5y-3z=8$. Then the points which lie on the plane $P=0$ is/are:

• A. $(0,9,17)$
• B. $(\frac{1}{7},2,\frac{1}{9})$
• C. $(1,3,-4)$
• D. $(\frac{1}{2},1,\frac{1}{3})$

Answer 1

Answer: (A), (D)

$(0,9,17)$
$(\frac{1}{2},1,\frac{1}{3})$

$P$ is the equation of the plane passing through the line of intersection of the planes $2x-y=0$ and $3z-y=0$.
i.e., $2x-y+\lambda (-y+3z)=0$
$\to 2x-(\lambda +1)y+3\lambda z=0$ ----- (1)
$P$ is perpendicular to $4x+5y-3z=0$
$\Rightarrow 8-5\lambda -5-9\lambda =0$
$\Rightarrow \lambda =\frac{3}{14}$
$\therefore$The equation of P is $28x-17y+9z=0$.
Clearly, $(0,9,17)$ and $(\frac{1}{2},1,\frac{1}{3})$ lie on the plane as they satisfy the equation of the plane.
Hence, options A and D are correct.