The magnitude of the projection of the vector $$\stackrel{\wedge }{2i}+\stackrel{\wedge }{3j}+\stackrel{\wedge }{k}$$ on the vector perpendicular to the plane containing the vectors $$\stackrel{\wedge }{i}+\stackrel{\wedge }{j}+\stackrel{\wedge }{k}$$ and $$\stackrel{\wedge }{i}+\stackrel{\wedge }{2j}+\stackrel{\wedge }{3k}$$ , is :

The vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y+ 4z = 5 which is perpendicular to the plane x – y + z = 0 is :

If a point R(4, y, z) lies on the line segment joining the points P(2, –3, 4) and Q(8, 0, 10), then the distance of R from the origin is :

A plane passing through the points (0, –1, 0) and (0, 0, 1) and making an angle {\pi \over 4} with the plane y – z + 5 = 0, also passes through the point

The equation of a plane containing the line of intersection of the planes 2x – y – 4 = 0 and y + 2z – 4 = 0 and passing through the point (1, 1, 0) is :

If the line, {{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 2} \over 4} meets the plane, x + 2y + 3z = 15 at a point P, then the distance of P from the origin is :

The vertices B and C of a $$\Delta$$ABC lie on the line, {{x + 2} \over 3} = {{y - 1} \over 0} = {z \over 4} such that BC = 5 units. Then the area (in sq. units) of this triangle, given that the point A(1, –1, 2), is :

Let P be the plane, which contains the line of intersection of the planes, x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to :

If the length of the perpendicular from the point ($$\beta$$, 0, $$\beta$$) ($$\beta$$ $$\ne$$ 0) to the line, {x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}} is \sqrt {{3 \over 2}}, then $$\beta$$ is equal to :

If Q(0, –1, –3) is the image of the point P in the plane 3x – y + 4z = 2 and R is the point (3, –1, –2), then the area (in sq. units) of $$\Delta$$PQR is :

If the plane 2x – y + 2z + 3 = 0 has the distances {1 \over 3} and {2 \over 3} units from the planes 4x – 2y + 4z + $$\lambda$$ = 0 and 2x – y + 2z + $$\mu$$ = 0, respectively, then the maximum value of $$\lambda$$ + $$\mu$$ is equal to :

A perpendicular is drawn from a point on the line {{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1} to the plane x + y + z = 3 such that the foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are :

If the line {{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 1} \over { - 1}} intersects the plane 2x + 3y – z + 13 = 0 at a point P and the plane 3x + y + 4z = 16 at a point Q, then PQ is equal to :

A plane which bisects the angle between the two given planes 2x – y + 2z – 4 = 0 and x + 2y + 2z – 2 = 0, passes through the point :

The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines $$\overrightarrow{r}=\left(\hat{i}+\hat{j}\right)+\lambda \left(\hat{i}+2\hat{j}-\hat{k}\right)$$ and $$\overrightarrow{r}=\left(\hat{i}+\hat{j}\right)+\mu \left(-\hat{i}+\hat{j}-2\hat{k}\right)$$ is :

Let S be the set of all real values of $$\lambda$$ such that a plane passing through the points (–$$\lambda$$2, 1, 1), (1, –$$\lambda$$2, 1) and (1, 1, – $$\lambda$$2) also passes through the point (–1, –1, 1). Then S is equal to :

The equation of the line passing through (–4, 3, 1), parallel to the plane x + 2y – z – 5 = 0 and intersecting the line {{x + 1} \over { - 3}} = {{y - 3} \over 2} = {{z - 2} \over { - 1}} is :

The plane through the intersection of the planes x + y + z = 1 and 2x + 3y – z + 4 = 0 and parallel to y-axis also passes through the point :

The equation of the plane containing the straight line {x \over 2} = {y \over 3} = {z \over 4} and perpendicular to the plane containing the straight lines {x \over 3} = {y \over 4} = {z \over 2} and {x \over 4} = {y \over 2} = {z \over 3} is :

If the lines x = ay + b, z = cy + d and x = a'z + b', y = c'z + d' are perpendicular, then :