Question

Consider two planes $P_1:2x-3y+z=-5$ and $P_2:x-4y-14z=5$. If a line $L$ which is perpendicular to line of intersection of planes $P_1\&P_2$ and whose one point $A\left(-7,y_1,z_1\right)$ lies on the plane $P_1$ and points $B\left(x_2,y_2,z_2\right)$ and $C\left(x_3,y_3,z_3\right)$ lie on plane $P_2$ has the equation $\frac{x+7}{a}=\frac{y-y_1}{-b}$ $=\frac{z-z_I}{c}$, where $a,b,c\in N$, then

• A. least value of $(a+b+c)$ is 6 .
• B. $y_1+z_1$ is equal to -3 .
• C. $\frac{z_3-z_2}{y_3-y_2}=3$
• D. $\frac{x_3-x_2}{z_3-z_2}=2$

Answer 1

Answer: (A), (B), (D)

Planes $P_1$ and $P_2$ are perpendicular
$L:\frac{x+7}{a}=\frac{y-y_1}{-b}=\frac{z-z_1}{c}$
The above line lies in plane $P_2$ and perpendicular to plane $P_1$
$\therefore A\left(-7,y_1,z_1\right)$ lies on both the planes.
$\therefore -14-3y_1+z_1=-5$ ....(i)
$-7-4y_1-14z_1=5$ ....(ii)

Solving (i) and (ii), we get $y_1=-3$ and $z_1=0$
$\therefore L:\frac{x+7}{2}=\frac{y+3}{-3}=\frac{z-0}{1}$