JEE MAIN 2026 Previous Year Questions (PYQs) – Page 24 of 25

$\text{Tr}(A) = 4 \Rightarrow \alpha + 2 = 4 \Rightarrow \alpha = 2$

$\text{Tr}(B) = 3 \Rightarrow \beta + 1 = 3 \Rightarrow \beta = 2$

$A^2 - 4A + 2I = 0$

$A^3 = 4A^2 - 2A = 16A - 8I - 2A = 14A - 8I$

$= \begin{bmatrix} 28 & 28 \ 14 & 20 \end{bmatrix}$

$B^2 - 3B + I = 0 \Rightarrow B^2 = 3B - I$

$B^3 = 3B^2 - B = 3(3B - I) - B = 8B - 3I$

$= \begin{bmatrix} 5 & 8 \ 8 & 13 \end{bmatrix}$

$A^3 - B^3 = \begin{bmatrix} 15 & 20 \ 6 & 7 \end{bmatrix}$

$\Rightarrow |A^3 - B^3| = 105 - 120 = -15$

$\Rightarrow \det(\text{adj}(A^3 - B^3)) = |A^3 - B^3|^2 = 225$

$6\int_{0}^{\pi/2} 2\sin 2x \cos x + \sin 2x , dx$

$= 6\int_{0}^{\pi/2} (4\sin x \cos^2 x + 2\sin x \cos x) dx$

$I = 12\int_{0}^{\pi/2} \sin x (2\cos^2 x + \cos x) dx$

Put $\cos x = t \Rightarrow -\sin x dx = dt$

$I = -12\int_{1}^{0} (2t^2 + t) dt$

$I = 12\int_{0}^{1} (2t^2 + t) dt$

$I = 12\left[\frac{2t^3}{3} + \frac{t^2}{2}\right]_0^1 = 12\left(\frac{2}{3} + \frac{1}{2}\right) = 17$

$\therefore$ parameter of point $A$ is $t = 2$

$\Rightarrow$ Parameter of point $B$ is $t = -\frac{1}{2}$

$\Rightarrow$ Coordinates of $B$ is $(1, -4)$

Case 1:

$A(16,16),; P(\alpha,\beta),; B(1,-4)$

$\alpha = \frac{5\cdot1 + 2\cdot16}{7} = \frac{37}{7}$

$\beta = \frac{5(-4) + 2\cdot16}{7} = \frac{12}{7}$

$\Rightarrow \alpha + \beta = 7$

Case 2:

$A(16,16),; P(\alpha,\beta),; B(1,-4)$

$\alpha = \frac{2\cdot1 + 5\cdot16}{7}$

$\beta = \frac{2(-4) + 5\cdot16}{7}$

$\Rightarrow \alpha + \beta = 22$

So minimum value of $\alpha + \beta = 7$

Equation of line is

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

$\therefore$ General point $R$ on line is $R(\lambda - 3, \lambda + 5, \lambda + 2)$

$P(-2, r, 1)$

$\vec{PR} = (\lambda - 1,; \lambda + 5 - r,; \lambda + 1)$

Now $\vec{PR} \cdot \vec{d} = 0$

$\Rightarrow (\lambda - 1) + (\lambda + 5 - r) + (\lambda + 1) = 0$

$\Rightarrow 3\lambda - r + 5 = 0$

$\Rightarrow \lambda = \frac{r - 5}{3}$

$\therefore R\left(\frac{r-5}{3} - 3,; \frac{r-5}{3} + 5,; \frac{r-5}{3} + 2\right)$

$= \left(\frac{r-14}{3},; \frac{r+10}{3},; \frac{r+1}{3}\right)$

Now

$PR = \sqrt{\frac{14}{3}}$

$\Rightarrow (PR)^2 = \frac{14}{3}$

$\Rightarrow \left(\frac{r-8}{3}\right)^2 + \left(\frac{10-2r}{3}\right)^2 + \left(\frac{r-2}{3}\right)^2 = \frac{14}{3}$

$\Rightarrow r^2 - 10r + 21 = 0$

$\Rightarrow r = 3, 7$

Sum of possible values of $r = 10$

$L_1 : \frac{x-2}{-3} = \frac{y-6}{2} = \frac{z-7}{4}$

Point $C$ on $L_1$: $(-3\lambda_1 + 2,; 2\lambda_1 + 6,; 4\lambda_1 + 7)$

$L_2 : \frac{x-4}{2} = \frac{y-3}{1} = \frac{z-5}{3}$

Point $D$ on $L_2$: $(2\lambda_2 + 4,; \lambda_2 + 3,; 3\lambda_2 + 5)$

D.R’s of line $L_3$:

$\frac{2\lambda_1 + 3\lambda_1 + 2}{-3} = \frac{\lambda_2 - 2\lambda_1 - 3}{5} = \frac{3\lambda_2 - 4\lambda_1 - 2}{16}$

$\lambda_1 = -3,; \lambda_2 = 2$

$C(11, 0, -5)$

$D(8, 5, 11)$

$|CD|^2 = 3^2 + 5^2 + 16^2 = 290$