Question

The equation of the plane passing through the intersection of the planes 2x – 5y + z = 3 and x + y + 4z = 5 and Parallel to the plane x + 3y + 6z = 1 is x + 3y + 6z = k, where k is

• A. 5
• B. 3
• C. 7
• D. 2

Answer 1

The correct option is B 3

Equation of plane passing through the intersection on the planes
$2x-5y+z=3$ and $x+y+4z=5$ is
$(2x-5y+z-3)+\lambda (x+y+4z)=0$
$\Rightarrow (2+\lambda )x+(-5+\lambda )y+(1+4\lambda )-3-5\lambda =0$ .........(i)
Which is parallel to the plane $x+3y+6z=1.$
Then, $\frac{2+\lambda }{1}=\frac{-5+\lambda }{3}=\frac{1+4\lambda }{6}$
$\therefore \lambda =\frac{-11}{2}$
From Eq.(i),
$-\frac{7}{2}x-\frac{21}{2}y-21z+\frac{49}{2}=0\therefore x+3y+6z=7$
Hence, $k=7$