The equation of the plane which contains the y-axis and passes through the point (1, 2, 3) is :

If the foot of the perpendicular from point (4, 3, 8) on the line $${L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}$$, l $$\ne$$ 0 is (3, 5, 7), then the shortest distance between the line L₁ and line $${L_2}:{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$$ is equal to :

If (x, y, z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of the expression
$$3 + {{x - 11} \over {{{(y - 19)}^2}{{(z - 12)}^2}}} + {{y - 19} \over {{{(x - 11)}^2}{{(z - 12)}^2}}} + {{z - 12} \over {{{(x - 11)}^2}{{(y - 19)}^2}}} - {{x + y + z} \over {14(x - 11)(y - 19)(z - 12)}}$$ is equal to :

Let the position vectors of two points P and Q be 3$$\widehat i$$ $$-$$ $$\widehat j$$ + 2$$\widehat k$$ and $$\widehat i$$ + 2$$\widehat j$$ $$-$$ 4$$\widehat k$$, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are (4, $$-$$1, 2) and ($$-$$2, 1, $$-$$2), respectively. Let lines PR and QS intersect at T. If the vector $$\overrightarrow {TA} $$ is perpendicular to both $$\overrightarrow {PR} $$ and $$\overrightarrow {QS} $$ and the length of vector $$\overrightarrow {TA} $$ is $$\sqrt 5 $$ units, then the modulus of a position vector of A is :