Q71 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics
A plane $\pi$ passing through the points $2 \hat{i}-3 \hat{j}, 3 \hat{i}+4 \hat{k}$ is parallel to the vector $2 \hat{i}+3 \hat{j}-4 \hat{k}$, If a line joining the points $\hat{i}+2 \hat{j}$ and $\hat{j}-2 \hat{k}$ intersects the plane $\pi$ at the point $a \hat{i}+b \hat{j}+c \hat{k}$, then $a+b+2 c=$
- Correct answerA$31$
- B$29$
- C$23$
- D$19$
Solution
Equation of plane $\pi$ is $a(x-2)+b(y+3)+c z=0$ $(3,0,4)$ lies on this plane
$a+3 b+4 c=0$ ...(i)
Normal to plane is perpendicular to $2 \hat{i}+3 \hat{j}-4 \hat{k}$.
$2 a+3 b-4 c=0$ ...(ii)
Solving (i) and (ii) we get
$a=-2 b, a=8 c, c=\frac{-b}{4}$
Equation of the plane is $-2 b(x-2)+b(y+3)-\frac{b}{4} z=0$ $\Rightarrow-2(x-2)+(y+3)-\frac{z}{4}=0 \Rightarrow 8 x-4 y+z=28$
Line joining $(\hat{i}+2 \hat{j})$ and $(\hat{j}-2 \hat{k})$ is $\begin{aligned} & \frac{x-1}{1}=\frac{y-z}{1}=\frac{z-0}{2}=k \\ & x=k+1, y=2 k+2, z=2 k \end{aligned}$
This point lies on the plane $8(k+1)-4(k+2)+2 k=28$ $\begin{aligned} & \Rightarrow 6 k=28 \Rightarrow k=\frac{14}{3} \Rightarrow a=\frac{17}{3}, b=\frac{20}{3}, c=\frac{28}{3} \\ & a+b+2 c=\frac{17+20+56}{3}=\frac{93}{3}=31 \end{aligned}$
Equation of the plane is $-2 b(x-2)+b(y+3)-\frac{b}{4} z=0$ $\Rightarrow-2(x-2)+(y+3)-\frac{z}{4}=0 \Rightarrow 8 x-4 y+z=28$
Line joining $(\hat{i}+2 \hat{j})$ and $(\hat{j}-2 \hat{k})$ is $\begin{aligned} & \frac{x-1}{1}=\frac{y-z}{1}=\frac{z-0}{2}=k \\ & x=k+1, y=2 k+2, z=2 k \end{aligned}$
This point lies on the plane $8(k+1)-4(k+2)+2 k=28$ $\begin{aligned} & \Rightarrow 6 k=28 \Rightarrow k=\frac{14}{3} \Rightarrow a=\frac{17}{3}, b=\frac{20}{3}, c=\frac{28}{3} \\ & a+b+2 c=\frac{17+20+56}{3}=\frac{93}{3}=31 \end{aligned}$
Hence, the correct answer is option: 'A'.