Three Dimension...

Find the angle between the lines $$\overrightarrow { r } =3i+2j-4k+\lambda \left( i+2j+2k \right) $$ and $$\overrightarrow { r } =\left( 5j-2k \right) +\mu \left( 3i+2j+6k \right) $$ 

The d.c's of the normal to the plane $$2x-y+2z+5=0$$ are 

Find the angle between the two lines having direction ratio $$(1,1,2)$$ and $$\left( \left( \sqrt { 3 } -1 \right) ,\left( -\sqrt { 3 } -1 \right) ,4 \right) $$.

The plane which passes through the point $$(-1, 0, -6)$$ and perpendicular to the line whose direction ratios is $$(6, 20, -1)$$ also passes through the point:

Equation of the plane through the mid-point of the join of $$A(4,5,-10)$$ and $$B(-1,2,1)$$ and perpendicular to $$AB$$ is

If $$(2, 3, -1)$$ is the foot of the perpendicular from $$(4, 2, 1)$$ to a plane, then the equation of that plane is $$ax+by+cz=d$$. Then $$a+d$$ is

lf $$\theta $$ is the angle  between two lines whose d.cs are $$l_{1},m_{\mathrm{1}},n_{\mathrm{1}}$$ and $$l_{2},m_{2},n_{2}$$, then

$$\displaystyle \dfrac{\Sigma(l_{1}+l_{2})^{2}}{4\cos^{2}(\dfrac{\theta}{2})}+\dfrac{\Sigma(l_{\mathrm{I}}-l_{2})^{2}}{4\sin^{2}(\dfrac{\theta}{2})}=$$

The plane $$2x+3y+kz-7=0$$ is parallel to the line whose direction ratios are $$(2, -3, 1)$$, then $$k=$$

If the foot of the perpendicular from $$(0,0,0)$$ to the plane is $$(1,2,2)$$, then the equation of the plane is

The d.r's of the line of intersection of the planes $$x+y+z-1$$ $$=0$$ and $$2x+3y+4z-7$$ $$=0$$ are