JEE 2023 — Mathematics PYQ

Equation of the line is  

$\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$

Let $B=(2,4,-3)$.

From the line, take  
$A(1,2,-5)$ and $P(1,-3,7)$.

So,
$\overrightarrow{AB}=(2-1)\hat{i}+(4-2)\hat{j}+(-3+5)\hat{k}<br>=\hat{i}+2\hat{j}+2\hat{k}$

Direction vector of the line,

$\hat{n}=<br>\left|\begin{array}{ccc}<br>\hat{i}& amp;\hat{j}& amp;\hat{k}\\ <br>1& amp;-3& amp;7\\ <br>1& amp;2& amp;2<br>\end{array}\right|$

$=(-6-14)\hat{i}-(2-7)\hat{j}+(2+3)\hat{k}$

$=-20\hat{i}+5\hat{j}+5\hat{k}<br>=-5(4\hat{i}-\hat{j}-\hat{k})$

$\therefore$ Equation of the plane is

$4(x-1)-(y-2)-(z+5)=0$

$\Rightarrow 4x-4-y+2-z-5=0$

$\Rightarrow 4x-y-z-7=0$

$\therefore$ Image of point $(-1,3,4)$ is $(\alpha ,\beta ,\gamma )$.

So,

$\frac{\alpha +1}{4}<br>=\frac{\beta -3}{-1}<br>=\frac{\gamma -4}{-1}<br>=\frac{2(-4-3-4-7)}{16+1+1}<br>=-2$

$\Rightarrow \alpha =7,\beta =1,\gamma =2$

So,

$\alpha +\beta +\gamma =10$