The curve $y=a x^3+b x^2+c x+5$ touches the X - axis at $(-2,0)$ and cuts the Y -axis at a point Q where its gradient is 3 , then values of $a, b, c$ respectively, are

A variable plane passes through the fixed point $(3,2,1)$ and meets $X, Y$ and $Z$ axes at points $A$, B and C respectively. A plane is drawn parallel to YZ - plane through A , a second plane is drawn parallel to ZX -plan through B , a third plane is drawn parallel to XY - plane through C . Then locus of the point of intersection of these three planes, is

The value of $\sin \left(\cos ^{-1}\left(-\frac{1}{3}\right)-\sin ^{-1}\left(\frac{1}{3}\right)\right)$ is

Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}$, and $\overline{\mathrm{c}}$ be three non-zero vectors such that no two of these are collinear. If the vector $\bar{a}+2 \bar{b}$ is collinear with $\bar{c}$ and $\bar{b}+3 \bar{c}$ is collinear with $\overline{\mathrm{a}}$, then $\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}+6 \overline{\mathrm{c}}$ equals