JEE 2023 — Mathematics PYQ

$$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$$

$$(x,y,z)\to (2\lambda +1,3\lambda +2,4\lambda +3)$$

$$\text{Plane }2x+y-3z=4$$

$$\vec{r}\cdot \vec{n}=p$$

$$\Rightarrow \vec{r}\cdot (2\hat{i}+\hat{j}-3\hat{k})=4$$

$$\overrightarrow{PQ}\cdot \vec{n}=0$$

$$\Rightarrow (2\lambda +1)\times 2+(3\lambda +1)\times 1+(4\lambda +3)\times (-3)=0$$

$$\Rightarrow \lambda =0$$

$$Q(2\lambda +1,3\lambda +2,4\lambda +3)$$

$$Q(1,2,3)$$

$$\text{Equation of line }\frac{x-0}{1}=\frac{y-1}{1}=\frac{z-2}{1}=\mu$$

$$\text{distance of the line term }(1,-9,2)$$

$$\overrightarrow{PQ}\cdot (\hat{i}+\hat{j}+\hat{k})=0$$

$$u=3$$

$$Q^{\prime }=(-3,-2,-1)$$

$$P{Q}^{\prime }=\sqrt{16+49+9}=\sqrt{74}$$