The coordinates of the point,where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}$ meets the plane $2x+4y-z=3$,are

The distance of the plane $3x + 4y + 5z + 19 = 0$ from the point $(1, -1, 1)$ measured along a line parallel to the line with direction ratios $2, 3, 1$ is

The reflection of the point $(-1, 3, 4)$ with respect to the plane $x - 2y = 0$ is .....

The equation of the plane passing through the line of intersection of the planes $x + y + z = 1$ and $2x + 3y - z + 4 = 0$ and parallel to the $x$-axis is:

Let the equation of the plane,that passes through the point $(1,4,-3)$ and contains the line of intersection of the planes $3x-2y+4z-7=0$ and $x+5y-2z+9=0$,be $\alpha x+\beta y+\gamma z+3=0$. Then $\alpha+\beta+\gamma$ is equal to:

Let $\pi_1$ be the plane determined by the vectors $\hat{i}+\hat{j}$ and $\hat{j}+\hat{k}$,and $\pi_2$ be the plane determined by the vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{j}-\hat{k}$. Let $\vec{a}$ be a vector parallel to the line of intersection of $\pi_1$ and $\pi_2$. If $|\vec{a}|=\sqrt{14}$,then $|\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})|=$