Math JEE Adv. 2025 paper 1

 Let's analyze the problem step-by-step to verify each statement.

---

Step 1: Find the Direction Vector of $L_1$

The line $L_1$ is the intersection of the planes:

$$2x+3y+z=4\text{and}x+2y+z=5.$$

To find its direction vector, compute the cross product of the normal vectors of the two planes:

$${\mathbf{n}}_1=(2,3,1),{\mathbf{n}}_2=(1,2,1)\mathrm{.}$$$$\mathbf{d}={\mathbf{n}}_1\times {\mathbf{n}}_2=\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 3& 1\\ 1& 2& 1\end{array}\mid =(3\cdot 1-1\cdot 2)\mathbf{i}-(2\cdot 1-1\cdot 1)\mathbf{j}+(2\cdot 2-3\cdot 1)\mathbf{k}=(1,-1,1)\mathrm{.}$$

Thus, the direction vector of $L_1$ is $\mathbf{d}=(1,-1,1)$.

---

Step 2: Parametric Equation of $L_2$

The line $L_2$ passes through $P(2,-1,3)$ and is parallel to $L_1$, so its parametric equations are:

$$L_2:\mathbf{r}=(2,-1,3)+t(1,-1,1)\mathrm{.}$$

---

Step 3: Find the Intersection Point $Q$ of $L_2$ with Plane $M$

The plane $M$ is given by:

$$2x+y-2z=6.$$

Substitute the parametric equations of $L_2$ into $M$:

$$2(2+t)+(-1-t)-2(3+t)=6.$$

Simplify:

$$4+2t-1-t-6-2t=6⟹-3-t=6⟹t=-9.$$

Thus, the coordinates of $Q$ are:

$$Q=(2+(-9),-1-(-9),3+(-9))=(-7,8,-6)\mathrm{.}$$

---

Step 4: Find the Foot of the Perpendicular $R$ from $P$ to $M$

The normal vector of $M$ is $\mathbf{n}=(2,1,-2)$. The parametric equations of the line through $P$ perpendicular to $M$ are:

$$\mathbf{r}=(2,-1,3)+s(2,1,-2)\mathrm{.}$$

Substitute into $M$:

$$2(2+2s)+(-1+s)-2(3-2s)=6.$$

Simplify:

$$4+4s-1+s-6+4s=6⟹-3+9s=6⟹s=1.$$

Thus, the coordinates of $R$ are:

$$R=(2+2(1),-1+1(1),3-2(1))=(4,0,1)\mathrm{.}$$

---

Step 5: Verify the Statements

(A) Length of $PQ$

$$PQ=\sqrt{(-7-2{)}^2+(8-(-1){)}^2+(-6-3{)}^2}=\sqrt{(-9{)}^2+9^2+(-9{)}^2}=\sqrt{81+81+81}=\sqrt{243}=9\sqrt{3}\mathrm{.}$$

This is true.

(B) Length of $QR$

$$QR=\sqrt{(4-(-7){)}^2+(0-8{)}^2+(1-(-6){)}^2}=\sqrt{{11}^2+(-8{)}^2+7^2}=\sqrt{121+64+49}=\sqrt{234}\mathrm{.}$$

This is false (not 15).

(C) Area of $\mathrm{△}PQR$

Compute the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$:

$$\overrightarrow{PQ}=(-9,9,-9),\overrightarrow{PR}=(2,1,-2)\mathrm{.}$$

The cross product is:

$$\overrightarrow{PQ}\times \overrightarrow{PR}=\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -9& 9& -9\\ 2& 1& -2\end{array}\mid =(-9,-36,-27)\mathrm{.}$$

The magnitude is:

$$\mathrm{\mid }\overrightarrow{PQ}\times \overrightarrow{PR}\mathrm{\mid }=\sqrt{(-9{)}^2+(-36{)}^2+(-27{)}^2}=\sqrt{81+1296+729}=\sqrt{2106}=3\sqrt{234}\mathrm{.}$$

The area of $\mathrm{△}PQR$ is:

$$\frac{1}{2}\mathrm{\mid }\overrightarrow{PQ}\times \overrightarrow{PR}\mathrm{\mid }=\frac{3}{2}\sqrt{234}\mathrm{.}$$

This is true.

(D) Angle Between $PQ$ and $PR$

The dot product is:

$$\overrightarrow{PQ}\cdot \overrightarrow{PR}=(-9)(2)+(9)(1)+(-9)(-2)=-18+9+18=9.$$

The magnitudes are:

$$\mathrm{\mid }\overrightarrow{PQ}\mathrm{\mid }=9\sqrt{3},\mathrm{\mid }\overrightarrow{PR}\mathrm{\mid }=\sqrt{2^2+1^2+(-2{)}^2}=3.$$

The cosine of the angle $\theta$ is:

$$\cos \theta =\frac{9}{9\sqrt{3}\cdot 3}=\frac{1}{3\sqrt{3}}\mathrm{.}$$

The acute angle is ${\cos }^{-1}\left(\frac{1}{3\sqrt{3}}\right)$, not ${\cos }^{-1}\left(\frac{1}{2\sqrt{3}}\right)$.
This is false.

---

Final Answer

The true statements are (A) and (C).

Answer: $\boxed{{\displaystyle A,C}}$