Three Dimension...

The three points $$ABC$$ have position vectors $$(1,x,3),(3,4,7)$$ and $$(y,-2,-5)$$ are collinear then $$(x,y)=$$

The points with position vectors $$\vec{a}+\vec{b},\vec{a}-\vec{b}$$ and $$\vec{a}+\lambda\vec{b}$$ are collinear for

If $$A = (1, 2, 3), B = (2, 10, 1)$$, $$Q$$ are collinear points and $$Q_x=-1$$, then $$Q_z=$$

If $$PQR$$ are the three points with respective position vectors $$\hat{i}+\hat{j},\ \hat{i}-\hat{j}$$ and $$a\hat{i}+b\hat{j}+c\hat{k}$$, then the points $$PQR$$ are collinear if

The three points whose position vectors are $$\overline{i}+2\overline{j}+3\overline{k,}$$ $$3\overline{i}+4\overline{j}+7\overline{k,}$$ and  $$-3\overline{i}-2\overline{j}-5\overline{k}$$

If $$\vec{a},\vec{b},\vec{c}$$ are the position vectors of points lie on a line, then $$\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}=$$

The product of the d.r's of a line perpendicular to the plane passing through the points $$(4,0,0),(0,2,0)$$ and $$( 1,0,1)$$ is

If $$R(a+2,a+3,a+4)$$ divides the line segment joining $$P(2, 3, 4)$$ and $$Q(4, 5, 6) $$ in the ratio $$-3:2$$, then the value of the parameter which represents $$a$$ is

lf the equation of the plane perpendicular to the $$\mathrm{z}$$ -axis and passing through the point $$(2, -3,4)$$ is $$ax+by+cz=d$$ then $$\displaystyle \dfrac{a+b+c}{d}=$$

The equation to the plane bisecting the line segment joining $$(-3, 3, 2), (9, 5, 4)$$ and perpendicular to the line segment is