Given that two vertex of a $\triangle A B C$ be $A(2,4,6)$ and

$B(0,-2,-5)$ and $G=$ centroid $=(2,1,-1)$ [Given]

Let, the other vertex is $C(x, y, z)$

According to the question,

$$ \begin{aligned} & \frac{2+0+x}{3}=2 \\\\ &\Rightarrow x =4 \end{aligned} $$

$$ \begin{aligned} &\frac{4-2+y}{3}=1 \\\\ &\Rightarrow y=1 \end{aligned} $$

$$ \begin{aligned} & \frac{6-5+z}{3}=-1 \\\\ &\Rightarrow z=1 \end{aligned} $$

Hence, third vertex is $C(4,1,-4)$

Now, if image of $C(4,1,-4)$ in the plane $x+2 y+4 z=11$ is $D(\alpha, \beta, \gamma)$.

So, $\frac{\alpha-4}{1}=\frac{\beta-1}{2}=\frac{\gamma+4}{4}=k$ (say)

JEE Main 2023 (Online) 10th April Morning Shift Mathematics - 3D Geometry Question 121 English Explanation 1

JEE Main 2023 (Online) 10th April Morning Shift Mathematics - 3D Geometry Question 121 English Explanation 1

$$ \Rightarrow \alpha=k+4, \beta=2 k+1, \gamma=4 k-4 $$

Then, co-ordinates of $D$ is $(k+4,2 k+1,4 k-4)$

Let, $E$ be the mid-point of $C D$, which lies on the plane $x+2 y+4 z=11$.

Then, co-ordinates of $E$ is

$$ \left(\frac{k+8}{2}, \frac{2 k+2}{2}, \frac{4 k-8}{2}\right)=\left(\frac{k+8}{2}, k+1,2 k-4\right) $$

Since, $E$ lies on the plane, $x+2 y+4 z-11=0$

$$ \begin{aligned} & \text { So, } \frac{k+8}{2}+2 k+2+8 k-16-11=0 \\\\ & \Rightarrow \frac{k+8+4 k+4+16 k-54}{2}=0 \end{aligned} $$

$$ \begin{array}{rlrl} & \Rightarrow 21 k =42 \\\\ & \Rightarrow k =2 \end{array} $$

Hence, $D \equiv(6,5,4) \equiv(\alpha, \beta, \gamma)$

JEE Main 2023 (Online) 10th April Morning Shift Mathematics - 3D Geometry Question 121 English Explanation 2

JEE Main 2023 (Online) 10th April Morning Shift Mathematics - 3D Geometry Question 121 English Explanation 2

So, $\alpha=6, \beta=5, \gamma=4$

Now, $\alpha \beta+\beta \gamma+\gamma \alpha=6 \times 5+5 \times 4+4 \times 6$

$=30+20+24=74$

3d Geometry

Continue this paper

Similar 3d Geometry PYQs

More practice from this question