Question

The plane containing the line $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and parallel to the line $\frac{x}{1}=\frac{y}{1}=\frac{z}{4}$ passes through the point :

• A. $(-1,-2,0)$
• B. $(1,0,5)$
• C. $(1,-2,5)$
• D. $(0,3,-5)$

Answer 1

The correct option is B $(1,0,5)$

Let the equation of the plane be $ax+by+cx+d=0$, where $a,b,c$ are direction ratios of normal to the plane.

The plane contains the line $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$.

Hence, it passes through the point $(1,2,3)$ and is parallel to the line.

$\Rightarrow a(x-1)+b(y-2)+c(z-3)=0$ and $a+2b+3c=0$

The given plane is parallel to $\frac{x}{1}=\frac{y}{1}=\frac{z}{4}$

$\Rightarrow a+b+4c=0$

Eliminating $a,b,c$ from the above equations gives

$\left|\begin{array}{ccc}x-1& y-2& z-3\\ 1& 2& 3\\ 1& 1& 4\end{array}\right|=0\Rightarrow 5x-y-z=0$

Clearly, $(1,0,5)$ lies on the plane $5x-y-z=0$

Hence, option B is correct.