Line $L_1$ passes through the point $(1,2,3)$ and is parallel to $z$-axis. Line $L_2$ passes through the point $(\lambda ,5,6)$ and is parallel to $y$-axis. Let for $\lambda=\lambda_1, \lambda_2, \lambda_2

Let A be the point of intersection of the lines ${\mathrm{L}}_1:\frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}$ and ${\mathrm{L}}_2:\frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$. Let B and C be the points on the lines ${\mathrm{L}}_1$ and ${\mathrm{L}}_2$ respectively such that $AB=AC=\sqrt{15}$. Then the square of the area of the triangle $A B C$ is :

Let the values of p , for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{\mathrm{r}}=(\mathrm{p}\hat{i}+2\hat{j}+\hat{k})+\lambda (2\hat{i}+3\hat{j}+4\hat{k})$ is $\frac{1}{\sqrt{6}}$, be $\mathrm{a}, \mathrm{b},(\mathrm{a}

Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha }{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta }{4}$ be $3\sqrt{30}$. Then the positive value of $5\alpha +\beta$ is

Let $A$ and $B$ be two distinct points on the line $L:\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2\sqrt{17}$ from the foot of perpendicular drawn from the point $(1,2,3)$ on the line $L$. If $O$ is the origin, then $\overrightarrow{OA}\cdot \overrightarrow{OB}$ is equal to

Each of the angles $\beta$ and $\gamma$ that a given line makes with the positive $y$ - and $z$-axes, respectively, is half of the angle that this line makes with the positive $x$-axes. Then the sum of all possible values of the angle $\beta$ is

The distance of the point $(7,10,11)$ from the line $\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$ along the line $\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}$ is

Let $x_1,x_2,...,x_{10}$ be ten observations such that $\sum _{i=1}^{10}(x_i-2)=30$, $\sum _{i=1}^{10}(x_i-\beta {)}^2=98$, $\beta >2$, and their variance is $\frac{4}{5}$. If $\mu$ and ${\sigma }^2$ are respectively the mean and the variance of $2(x_1-1)+4\beta ,2(x_2-1)+4\beta ,...,2(x_{10}-1)+4\beta$, then $\frac{\beta \mu }{{\sigma }^2}$ is equal to :

Marks obtains by all the students of class 12 are presented in a freqency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18 , then the total number of students is :

For a statistical data ${\mathrm{x}}_1,{\mathrm{x}}_2,\dots ,{\mathrm{x}}_{10}$ of 10 values, a student obtained the mean as 5.5 and ${\sum }_{i=1}^{10}{x}_i^2=371$. He later found that he had noted two values in the data incorrectly as 4 and 5 , instead of the correct values 6 and 8 , respectively. The variance of the corrected data is

The mean and standard deviation of 100 observations are 40 and 5.1 , respectively. By mistake one observation is taken as 50 instead of 40 . If the correct mean and the correct standard deviation are $\mu$ and $\sigma$ respectively, then $10(\mu +\sigma )$ is equal to

If the mean and the variance of $6,4, a, 8, b, 12,10,13$ are 9 and 9.25 respectively, then $a+b+a b$ is equal to :

Let the mean and the standard deviation of the observation $2,3,3,4,5,7, a, b$ be 4 and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is :

Let the Mean and Variance of five observations $x_1=1,x_2=3,x_3=a,x_4=7$ and $x_5=\mathrm{b},a>\mathrm{b}$, be 5 and 10 respectively. Then the Variance of the observations $n+x_n,n=1,2,\dots ,5$ is

Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is :

Let $a_1,a_2,a_3,\dots$ be a G.P. of increasing positive terms. If $a_1{a}_5=28$ and $a_2+a_4=29$, then $a_6$ is equal to:

Suppose that the number of terms in an A.P. is $2k,k\in N$. If the sum of all odd terms of the A.P. is 40 , the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27 , then k is equal to:

For positive integers $n$, if $4a_n=\left(n^2+5n+6\right)$ and $S_n=\sum _{k=1}^n\left(\frac{1}{a_k}\right)$, then the value of $507S_{2025}$ is :

If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to

Let $S_n=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\dots$ upto $n$ terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is $\sqrt{2026{\mathrm{S}}_{2025}}$, then the absolute difference betwen $20^{\text{th }}$ and $15^{\text{th }}$ terms of the A.P. is