Let $$a, b, c$$ be three distinct real numbers, none equal to one. If the vectors $$a \hat{i}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{i}}+b \hat{j}+\hat{\mathrm{k}}$$ and $$\hat{\mathrm{i}}+\hat{\mathrm{j}}+c \hat{\mathrm{k}}$$ are coplanar, then $$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$ is equal to :
Let $$\lambda \in \mathbb{Z}, \vec{a}=\lambda \hat{i}+\hat{j}-\hat{k}$$ and $$\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$$. Let $$\vec{c}$$ be a vector such that $$(\vec{a}+\vec{b}+\vec{c}) \times \vec{c}=\overrightarrow{0}, \vec{a} \cdot \vec{c}=-17$$ and $$\vec{b} \cdot \vec{c}=-20$$. Then $$|\vec{c} \times(\lambda \hat{i}+\hat{j}+\hat{k})|^{2}$$ is equal to :
If the point $$\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$$ lies on the curve traced by the mid-points of the line segments of the lines $$x \cos \theta+y \sin \theta=7, \theta \in\left(0, \frac{\pi}{2}\right)$$ between the co-ordinates axes, then $$\alpha$$ is equal to :
Let $$\alpha, \beta$$ be the roots of the quadratic equation $$x^{2}+\sqrt{6} x+3=0$$. Then $$\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}$$ is equal to :