If the distances of the point $(1,2, a)$ from the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $\mathrm{L}_1: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $\mathrm{L}_2: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equal, then $a+b+c$ is equal to

For three unit vectors $\vec{a}, \vec{b}, \vec{c}$ satisfying

$$ |\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9 \text { and }|2 \vec{a}+k \vec{b}+k \vec{c}|=3 \text {, } $$

the positive value of k is :

For some $\theta \in\left(0, \frac{\pi}{2}\right)$, let the eccentricity and the length of the latus rectum of the hyperbola $x^2-y^2 \sec ^2 \theta=8$ be $e_1$ and $l_1$, respectively, and let the eccentricity and the length of the latus rectum of the ellipse $x^2 \sec ^2 \theta+y^2=6$ be $e_2$ and $l_2$, respectively. If $e_1^2=e_2^2\left(\sec ^2 \theta+1\right)$, then $\left(\frac{l_1 l_2}{e_1 e_2}\right) \tan ^2 \theta$ is equal to

In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is $\mathbb{R}-(a, b)$, then $a^2+b^2$ is equal to

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