NTA JEE Main 13th April 2023 Shift 2

If the system of equations
$$2x + y - z = 5$$
$$2x - 5y + \lambda z = \mu$$
$$x + 2y - 5z = 7$$
has infinitely many solutions, then $$(\lambda + \mu)^2 + (\lambda - \mu)^2$$ is equal to

The range of $$f(x) = 4\sin^{-1}\left(\frac{x^2}{x^2+1}\right)$$ is

The value of $$\frac{e^{-\frac{\pi}{4}} + \int_0^{\frac{\pi}{4}} e^{-x}\tan^{50}x \ dx}{\int_0^{\frac{\pi}{4}} e^{-x}(\tan^{49}x + \tan^{51}x) \ dx}$$

The area of the region $$\{x, y : x^2 \leq y \leq x^2 - 4, y \geq 1\}$$ is

Let $$|\vec{a}| = 2$$, $$|\vec{b}| = 3$$ and the angle between the vectors $$\vec{a}$$ and $$\vec{b}$$ be $$\frac{\pi}{4}$$. Then $$|(\vec{a} + 2\vec{b}) \times (2\vec{a} - 3\vec{b})|^2$$ is equal to

Let for a triangle $$ABC$$
$$\vec{AB} = -2\hat{i} + \hat{j} + 3\hat{k}$$
$$\vec{CB} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$$
$$\vec{CA} = 4\hat{i} + 3\hat{j} + \delta\hat{k}$$
If $$\delta > 0$$ and the area of the triangle $$ABC$$ is $$5\sqrt{6}$$ then $$\vec{CB} \cdot \vec{CA}$$ is equal to

The plane, passing through the points $$(0, -1, 2)$$ and $$(-1, 2, 1)$$ and parallel to the line passing through $$(5, 1, -7)$$ and $$(1, -1, -1)$$, also passes through the point

The line, that is coplanar to the line $$\frac{x+3}{-3} = \frac{y-1}{1} = \frac{z-5}{5}$$, is

Let N be the foot of perpendicular from the point $$P(1, -2, 3)$$ on the line passing through the points $$(4, 5, 8)$$ and $$(1, -7, 5)$$. Then the distance of N from the plane $$2x - 2y + z + 5 = 0$$ is

The random variable $$X$$ follows binomial distribution $$B(n, p)$$, for which the difference of the mean and the variance is 1.
If $$2P(X = 2) = 3P(X = 1)$$, then $$n^2P(X > 1)$$ is equal to