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

The coordinates of the point where the line through the points $$A(5,1,6)$$ and $$B(3,4,1)$$ crosses the yz-plane is

The point on the line $$\dfrac {x - 2}{1} = \dfrac {y + 3}{-2} = \dfrac {z + 5}{-2}$$ at a distance of 6 from the point $$\left ( 2,-3,-5 \right )$$ is

The distance between two points P and Q is d and the length of their projections of PQ on the coordinate planes are $$d_1,d_2,d_3$$. Then $$d_1^2 +d_2^2 + d_3^2 = kd^2$$ where K is

The shortest distance between the lines $$\dfrac {x - 3}{2} = \dfrac {y + 15}{-7} = \dfrac {z - 9}{5}$$ and $$\dfrac {x + 1}{2} = \dfrac {y - 1}{1} = \dfrac {z - 9}{-3}$$ is

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

The point on the line $$\frac{x - 2} {1} = \frac{y + 3} {-2} = \frac{z + 5} {-2} $$ at a distance of 6 from the point $$\left ( 2, -3, -5 \right )$$ is

The points $$A(1,2,-1),B(2,5,-2),C(4,4,-3)$$ and $$D(3,1,-2)$$ are

Let $$A \left ( 2\hat{i}+3\hat{j}+5\hat{k} \right )B\left ( -\hat{i}+3\hat{j}+2\hat{k} \right ) $$and $$ C \left ( \lambda \hat{i}+5\hat{j}+\mu\hat{k} \right )$$ are vertices of a triangle and its median through $$A$$ is equally inclined to the positive directions of the axes. The value of $$\lambda+\mu$$ is equal to

If $$(0, b, 0)$$ is the centroid of the triangle formed by the points $$(4, 2, -3)$$ , $$({a}, -5, 1)$$ and $$(2, -6, 2)$$ . If $$a ,b$$ are the roots of the quadratic equation $$ x^2+px+q = 0 $$, then $$p,q$$ are