The angle between the line $\frac{x + 1}{3} = \frac{y - 1}{4} = \frac{z - 2}{2}$ and the plane $2x - 3y + z + 4 = 0$ is:

If the line passing through the points $(a, 2, -4)$ and $(5, 3, b)$ crosses the $ZX$-plane at the point $(-a+2b, 0, a+b)$,then find the value of $14a+7b$.

$A$ line $L$ is parallel to both the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If line $L$ makes an angle $\alpha$ with the positive direction of the $X$-axis,then $\cos \alpha =$

The coordinates of the point where the line joining the points $(2, -3, 1)$ and $(3, -4, -5)$ intersects the plane $2x + y + z = 7$ are:

The position vector of the point of intersection of the line $\bar{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$ and the $XOY$-plane is:

If $\lambda_1 < \lambda_2$ are two values of $\lambda$ such that the angle between the planes $P_1: \vec{r} \cdot (3 \hat{i} - 5 \hat{j} + \hat{k}) = 7$ and $P_2: \vec{r} \cdot (\lambda \hat{i} + \hat{j} - 3 \hat{k}) = 9$ is $\sin^{-1}\left(\frac{2 \sqrt{6}}{5}\right)$,then the square of the length of the perpendicular from the point $(38 \lambda_1, 10 \lambda_2, 2)$ to the plane $P_1$ is $...........$.