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 C 7

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