Let $$5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x > 0$$. Then $$18 \int_\limits{1}^{2} f(x) d x$$ is equal to :

Let $$I(x)=\int \frac{x^{2}\left(x \sec ^{2} x+\tan x\right)}{(x \tan x+1)^{2}} d x$$. If $$I(0)=0$$, then $$I\left(\frac{\pi}{4}\right)$$ is equal to :

Let the position vectors of the points A, B, C and D be $$5 \hat{i}+5 \hat{j}+2 \lambda \hat{k}, \hat{i}+2 \hat{j}+3 \hat{k},-2 \hat{i}+\lambda \hat{j}+4 \hat{k}$$ and $$-\hat{i}+5 \hat{j}+6 \hat{k}$$. Let the set $$S=\{\lambda \in \mathbb{R}$$ : the points A, B, C and D are coplanar $$\}$$.

Then $$\sum_\limits{\lambda \in S}(\lambda+2)^{2}$$ is equal to :

Let $$\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}-2 \hat{k}$$ and $$\vec{c}=-\hat{i}+4 \hat{j}+3 \hat{k}$$. If $$\vec{d}$$ is a vector perpendicular to both $$\vec{b}$$ and $$\vec{c}$$, and $$\vec{a} \cdot \vec{d}=18$$, then $$|\vec{a} \times \vec{d}|^{2}$$ is equal to :