Let $\mathbf{P}$ be a point on the line $L_1: \frac{x-2}{2}=y+1=\frac{z-1}{2}$ such that its distance from the point $A(2,-1,1)$ is 6 units.
Given that $\boldsymbol{x}$-coordinate of $\mathbf{P}$ is greater than $\mathbf{2}$,
Find the coordinates of point Q on the line $L_2: x-1=\frac{y-2}{2}=\frac{z-2}{2}$ such that $\mathbf{Q}$ is the closest point to $\mathbf{P}$.
$$ \left(-\frac{14}{9},-\frac{28}{9},-\frac{28}{9}\right) $$
$$ (2,4,4) $$
$$ (6,1,5) $$
$$ (1,2,2) $$
The range of the function $f(x)={ }^{(7-x)} P_{(x-3)}$ is
$\{1,2,3,4\}$
$\{1,2,3,4,5\}$
$\{1,2,3,4,5,6\}$
$\{1,2,3\}$
The set expression $A \cup\left(B \cap\left(A^{\prime} \cup B^{\prime}\right)\right)$ is equivalent to
$\left(A^{\prime} \cup B^{\prime}\right)^{\prime}$
$\xi$ (Universal set)
$A \cup B$
$A \cap B^{\prime}$
\text { The remainder when } \mathbf{7}^{\mathbf{1 0 3}} \text { is divided by } \mathbf{2 5} \text { is }
7
18
1
-1
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