The distance of the point having position vector $$\hat{i}-2 \hat{j}-6 \hat{k}$$, from the straight line passing through the point $$(2,-3,-4)$$ and parallel to the vector $$6 \hat{i}+3 \hat{j}-4 \hat{k}$$ is units.
A vector $$\overrightarrow{\mathrm{n}}$$ is inclined to $$\mathrm{X}$$-axis at $$45^{\circ}$$, $$\mathrm{Y}$$-axis at $$60^{\circ}$$ and at an acute angle to Z-axis If $$\overrightarrow{\mathrm{n}}$$ is normal to a plane passing through the point $$(-\sqrt{2}, 1,1)$$, then equation of the plane is
If $$\cos ^{-1} x-\cos ^{-1} \frac{y}{3}=\alpha$$, where $$-1 \leq x \leq 1, -3 \leq y \leq 3, x \leq \frac{y}{3}$$, then for all $$x, y$$ $$9 x^2-6 x y \cos \alpha+y^2$$ is equal to
If $$\mathrm{f}(1)=1, \mathrm{f}^{\prime}(1)=3$$, then the derivative of $$\mathrm{f}(\mathrm{f}(\mathrm{f}(x)))+(\mathrm{f}(x))^2$$ at $$x=1$$ is