Question

For the line $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$, which one of the following is incorrect?

• A. It lies in the plane $x-2y+z=0$
• B. It is same as line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$
• C. It passes through $(2,3,5)$
• D. It is parallel to the plane $x-2y+z-6=0$

Answer 1

Answer: (C)

It passes through $(2,3,5)$

Consider, $L:\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$
a) For $P:x-2y+z=0$
Since, $(1,2,3)$ lies on P
and $(i+2j+3k).(i-2j+k)=0$
Therefore, $L$ lies in $P$ and option A is correct
b) Since, $(1,2,3)$ satisfy the line $L_1:\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$
Therefore, $L$ & $L_1$ are same and option B is correct
c) $(2,3,5)$ does not satisfy $L:\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$
Therefore, option C is incorrect.
d) Since, $L$ is parallel to $P:x-2y+z=0$
It is also parallel to $x-2y+z-6=0$.
Therefore, option D is correct.
Ans: C