The line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 1}{4}$ meets the plane $x + 2y + 3z = 14$ at the point:

The value of $k$ such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies on the plane $2x-4y+z=7$ is

The length and foot of the perpendicular from the point $(7, 14, 5)$ to the plane $2x + 4y - z = 2$ are

The ratio in which the plane $\bar{r} \cdot (\hat{i}-2 \hat{j}+3 \hat{k})=17$ divides the line joining the points $-2 \hat{i}+4 \hat{j}+7 \hat{k}$ and $3 \hat{i}-5 \hat{j}+8 \hat{k}$ is:

Let $\pi_1$ be a plane passing through the point $\hat{i}+\hat{j}+\hat{k}$ and perpendicular to the vector $-\hat{j}+2\hat{k}$. Let the line $L$ passing through the points $3\hat{i}-2\hat{j}+\hat{k}$ and $-\hat{i}+3\hat{j}+\hat{k}$ be a normal to the plane $\pi_2$. If the angle between the planes $\pi_1$ and $\pi_2$ is $\theta$,then $\cos \theta =$

The position vectors of the points $A$ and $B$ are respectively $\hat{i}+2 \hat{j}$ and $2 \hat{i}+\hat{j}+\hat{k}$. If the points $P$ and $Q$ are respectively the orthogonal projections of $A$ and $B$ on the plane $x+y+z=3$,then $P Q=$