Question

The equation of the plane parallel to the lines x − 1 = 2y − 5 = 2z and 3x = 4y − 11 = 3z − 4 and passing through the point (2, 3, 3) is
(a) x − 4y + 2z + 4 = 0
(b) x + 4y + 2z + 4 = 0
(c) x − 4y + 2z − 4 = 0
(d) None of these

Answer 1

a) x − 4y + 2z + 4 = 0

$$\begin{gathered} \text{Let}a,b,c\text{be the direction ratios of the required plane.} \\ \text{The given line equations can be rewritten as} \\ \frac{x-1}{1}=\frac{y-{\displaystyle \frac{5}{2}}}{{\displaystyle \frac{1}{2}}}=\frac{z-0}{{\displaystyle \frac{1}{2}}}...\left(1\right) \\ \frac{x-0}{{\displaystyle \frac{1}{3}}}=\frac{y-{\displaystyle \frac{11}{4}}}{{\displaystyle \frac{1}{4}}}=\frac{z-{\displaystyle \frac{4}{3}}}{{\displaystyle \frac{1}{3}}}...\left(2\right) \\ \text{Since the required plane is parallel to the lines (1) and (2),} \\ a+\frac{b}{2}+\frac{c}{2}=0\Rightarrow 2a+b+c=0...\left(3\right) \\ \frac{a}{3}+\frac{b}{4}+\frac{c}{3}=0\Rightarrow 4a+3b+4c=0...\left(4\right) \\ \text{Solving (3) and (4) using cross-multiplication method, we get} \\ \frac{a}{1}=\frac{b}{-4}=\frac{c}{2}=\lambda \text{(say)} \\ \Rightarrow a=\lambda ,b=-4\lambda ,c=2\lambda \\ \text{Now, the equation of the plane whose direction ratios are}\lambda \text{, -4}\lambda ,2\lambda \text{and passing through the point (2, 3, 3) is} \\ \lambda \left(x-2\right)+\left(-4\lambda \right)\left(y-3\right)+2\lambda \left(z-3\right)=0 \\ \Rightarrow x-4y+2z+4=0 \end{gathered}$$
So, the answer is (a).