If the line $$\frac{1-x}{-3}=y=\frac{z+2}{2}$$ is perpendicular to the line $$\frac{3 x-1}{2 b}=3-y=\frac{z-1}{a}$$, then find the value of $$3 a+3 b$$

The lines $$\vec{r}=(2 \hat{\jmath}-3 \hat{k})+\lambda(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})$$ and $$\vec{r}=(2 \hat{\imath}+6 \hat{\jmath}+3 \hat{k})+\mu(2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})$$ are

A line makes the same angle $$\theta$$ with each of the $$x$$ and $$z$$-axes. If the angle $$\beta$$, which it makes with the $$y$$-axis is such that $$\sin ^2 \beta=3 \sin ^2 \theta$$, then $$\cos ^2 \theta$$ equals

The foot of the perpendicular from $$(2,4,-1)$$ to the line $$x+5=\frac{1}{4}(y+3)=-\frac{1}{9}(z-6)$$ is