Three Dimension...

If $$\overline{a}$$ and $$\overline{b}$$ are two non-collinear vectors, then the points $$l_{1}\overline{a}+m_{1}\overline{b}$$, $$  l_{2}\overline{a}+m_{2}\overline{b}$$ and $$l_{3}\overline{a}+m_{3}\overline{b}$$ are collinear if

A line makes an angle $$\alpha,\beta,\gamma$$ with the $$X,Y,Z$$ axes. Then $$\sin^2\alpha+\sin^2\beta+\sin^2\gamma=$$

The angle made by line $$r[cos \theta  - \sqrt{3} sin \theta ]=5$$ with initial line is

The line passing through the points $$10\hat{i}+3\hat{j}$$, $$ 12\hat{i}+5\hat{j}$$ also passes through the point $$a\hat{i}+11 \hat{j}$$, then $$a=$$

The position vectors of three points are $$2\overrightarrow { a } -\overrightarrow { b } +3\overrightarrow { c } ,\overrightarrow { a } -2\overrightarrow { b } +\lambda \overrightarrow { c } $$ and $$\mu \overrightarrow { a } -5\overrightarrow { b } $$, where $$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $$ are non-coplanar vectors. The points are coliinear when

The vectors $$2\hat{i}+3\hat{j};5\hat{i}+6\hat{j};8\hat{i}+\lambda\hat{j}$$ have their initial points at $$(1,1 )$$. The value of  $$\lambda$$ so that the vectors terminate on one straight line is

The point collinear with $$(1, -2, -3)$$ and $$(2, 0, 0)$$ among the following is

If the points $$(h, 3, -4), (0, -7, 10)$$ and $$(1, k, 3)$$ are collinear, then $$h + k$$ is

Assertion ($$A$$): The points with position vectors $$\overline{a},\overline{b},\overline{c}$$ are collinear if $$2\overline{a}-7\overline{b}+5\overline{c}=0$$.
Reason ($$R$$): The points with position vectors $$\overline{a},\overline{b},\overline{c}$$ are collinear if $$l\overline{a}+m\overline{b}+n\overline{c}=\overline{0}$$.

If the points whose position vectors are $$2\overline{i}+\overline{j}+\overline{k},\ 6\overline{i}-\overline{j}+2\overline{k}$$ and $$14\overline{i}-5\overline{j}+p\overline{k}$$ are collinear then the value of $$\mathrm{p}$$ is