Question
The variable plane $$\left( {2\lambda + 1} \right)x + \left( {3 - \lambda } \right)y + z = 4$$ always passes through the line :
A.
$$\frac{x}{0} = \frac{y}{0} = \frac{{z + 4}}{1}$$
B.
$$\frac{x}{1} = \frac{y}{2} = \frac{z}{{ - 3}}$$
C.
$$\frac{x}{1} = \frac{y}{2} = \frac{{z - 4}}{{ - 7}}$$
D.
none of these
Answer :
$$\frac{x}{1} = \frac{y}{2} = \frac{{z - 4}}{{ - 7}}$$
Solution :
The plane is $$x + 3y + z - 4 + \lambda \left( {2x - y} \right) = 0.$$
This always passes through the intersection of the planes $$x + 3y + z - 4 = 0$$ and $$2x - y = 0,$$ which is a line.
$$\eqalign{
& {\text{Now, }}2x - y = 0\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \frac{x}{1} = \frac{y}{2} \cr
& \therefore \,\,x + 3y + z - 4 = 0 \cr
& \Rightarrow x + 3.2x + z - 4 = 0 \cr
& \Rightarrow \,7x + z - 4 = 0 \cr
& \therefore \,7x = - \left( {z - 4} \right) \cr
& \Rightarrow \frac{x}{1} = \frac{{z - 4}}{{ - 7}} \cr
& \therefore \,\,{\text{ the line is }}\frac{x}{1} = \frac{y}{2} = \frac{{z - 4}}{{ - 7}}. \cr} $$