Let $$P\left( {\alpha ,\,\beta ,\,\gamma } \right)$$   be the foot of the perpendicular from the origin $$O\left( {0,\,0,\,0} \right)$$   to the plane So, the plane passes through $$P\left( {\alpha ,\,\beta ,\,\gamma } \right)$$   and is perpendicular to $$OP.$$  Clearly direction ratios of $$OP$$  i.e., normal to the plane are $${\alpha ,\,\beta ,\,\gamma }.$$   Therefore, equation of the plane is
$$\alpha \left( {x - \alpha } \right) + \beta \left( {y - \beta } \right) + \gamma \left( {z - \gamma } \right) = 0$$
This plane passes through the fixed point $$\left( {1,\,2,\,3} \right),$$   so
$$\eqalign{ & \alpha \left( {1 - \alpha } \right) + \beta \left( {2 - \beta } \right) + \gamma \left( {3 - \gamma } \right) = 0 \cr & {\text{or, }}{\alpha ^2} + {\beta ^2} + {\gamma ^2} - \alpha - 2\beta - 3\gamma = 0 \cr} $$
Generalizing $${\alpha ,\,\beta }$$  and $$\gamma $$, locus of $$P\left( {\alpha ,\,\beta ,\,\gamma } \right)$$   is
$${x^2} + {y^2} + {z^2} - x - 2y - 3z = 0$$

Since, equation of plane through intersection of planes
$$x+y+z=1$$   and $$2x+3y-z+4=0$$     is
$$\eqalign{ & \left( {2x + 3y - z + 4} \right) + \lambda \left( {x + y + z - 1} \right) = 0 \cr & \left( {2 + \lambda } \right)x + \left( {3 + \lambda } \right)y + \left( { - 1 + \lambda } \right)z + \left( {4 - \lambda } \right) = 0.....(1) \cr} $$
But, the above plane is parallel to y-axis then
$$\eqalign{ & \left( {2 + \lambda } \right) \times 0 + \left( {3 + \lambda } \right) \times 1 + \left( { - 1 + \lambda } \right) \times 0 = 0 \cr & \Rightarrow \lambda = - 3 \cr} $$
Hence, the equation of required plane is
$$\eqalign{ & - x - 4z + 7 = 0 \cr & \Rightarrow x + 4z - 7 = 0 \cr} $$
Therefore, (3, 2, 1) the passes through the point.

The planes are $$2x + y + 2z - 8 = 0.....(1)$$
and $$4x + 2y + 4z + 5 = 0$$
or $$2x + y + 2z + \frac{5}{2} = 0.....(2)$$
$$\therefore $$ Distance between (1) and (2)
$$ = \left| {\frac{{\frac{5}{2} + 8}}{{\sqrt {{2^2} + {1^2} + {2^2}} }}} \right| = \left| {\frac{{21}}{{2\sqrt 9 }}} \right| = \frac{7}{2}$$

$$\eqalign{ & {\text{The length of the edges are given by}} \cr & a = 5 - 2 = 3, \cr & b = 9 - 3 = 6, \cr & c = 7 - 5 = 2 \cr & {\text{So length of the diagonal is}} \cr & \sqrt {{a^2} + {b^2} + {c^2}} = \sqrt {9 + 36 + 4} = 7{\text{ units}} \cr} $$

Plane containing two lines $$\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$$   and $$\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$$   is given by
\[\left| \begin{array}{l} x\,\,\,\,\,y\,\,\,\,\,z\\ 3\,\,\,\,\,4\,\,\,\,\,2\\ 4\,\,\,\,\,2\,\,\,\,\,3 \end{array} \right| = 0\,\, \Rightarrow 8x - y - 10z = 0\]
Now equation of plane containing the line $$\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$$   and perpendicular to the plane $$8x-y-10z=0$$    is
\[\begin{array}{l} \left| \begin{array}{l} x\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,z\\ 2\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,4\\ 8\,\,\,\, - 1\,\,\,\, - 10 \end{array} \right| = 0\,\\ \, \Rightarrow - 26x + 52y - 26z = 0\,\,{\rm{ or }}\,\,x - 2y + z = 0 \end{array}\]

Equation of plane passing through $$\left( {1,\, - 3,\,1} \right)$$   and whose normal $$\left( {1,\, - 3,\,1} \right)$$   is
$$\eqalign{ & 1\left( {x - 1} \right) - 3\left( {y + 3} \right) + 1\left( {z - 1} \right) = 0 \cr & \Rightarrow x - 3y + z - 11 = 0 \cr & \Rightarrow \frac{x}{{11}} - \frac{y}{{\frac{{11}}{3}}} + \frac{z}{{11}} = 0 \cr} $$
The above plane intercept the $$x$$-axis at $$11.$$

The line has +ve and equal direction cosines, these are $$\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}$$   or direction ratios are 1, 1, 1. Also the lines passes through $$P\left( {2,\, - 1,\,2} \right).$$
$$\therefore $$ Equation of line is $$\frac{{x - 2}}{1} = \frac{{y + 1}}{1} = \frac{{z - 2}}{1} = \lambda \,\,\left( {{\text{say}}} \right)$$
Let $$Q\left( {\lambda + 2,\,\lambda - 1,\,\lambda + 2} \right)$$     be a point on this line where it meets the plane $$2x+y+z=9$$
Then $$Q$$ must satisfy the equation of plane
$${\text{i}}{\text{.e}}{\text{.,}}\,\,\,2\left( {\lambda + 2} \right) + \lambda - 1 + \lambda + 2 = 9\,\, \Rightarrow \lambda = 1$$
$$\therefore \,\,Q$$   has coordintes $$\left( {3,\,0,\,3} \right)$$
Hence the length of line segments $$PQ$$
$$ = \sqrt {{{\left( {2 - 3} \right)}^2} + {{\left( { - 1 - 0} \right)}^2} + {{\left( {2 - 3} \right)}^2}} = \sqrt 3 $$

Let equation of plane through $$z$$-axis is $$ax + by = 0$$
It is given that this plane is parallel to the line $$\frac{{x - 1}}{{\cos \,\theta }} = \frac{{y + 2}}{{\sin \,\theta }} = \frac{{z - 3}}{0}$$
Since the plane parallel to the line
$$\eqalign{ & \therefore \,a\,\cos \,\theta + b\,\sin \,\theta = 0 \cr & \Rightarrow a\,\cos \,\theta = - b\,\sin \,\theta \cr & \Rightarrow a = - b\,\tan \,\theta \cr & \therefore \, - b\,\tan \,\theta \,x + by = 0 \cr & \Rightarrow x\,\tan \,\theta - y = 0\,\,\,\,\,\,\,\left( {\because \,b \ne 0} \right) \cr} $$
which is required equation of plane.

Let there be cube of side $$'a'.$$  Co-ordinates of its vertices $$O,\,A,\,B,\,C,\,D,\,E,\,F$$     have be marked in the figure. Diagonals are $$OE,\,FC,\,GB$$    and $$AD.$$
Direction ratios $$\left( {d{r_3}} \right)$$  of these diagonals are :
$$\eqalign{ & OE\,\left\langle {\left( {a - 0} \right),\,\left( {a - 0} \right),\,\left( {a - 0} \right)} \right\rangle = \left( {a,\,a,\,a} \right)\,; \cr & FC\,\left\langle {\left( { - a,\,a,\, - a} \right)} \right\rangle \,; \cr & GB\,\left\langle {\left( { - a,\,a,\,a} \right)} \right\rangle {\text{ and }}AD\,\left\langle {\left( {a,\,a,\, - a} \right)} \right\rangle \cr & {\text{Their dcs are :}} \cr & OE,\,\left\langle {\frac{a}{{\sqrt {{a^2} + {a^2} + {a^2}} }},\,\frac{a}{{\sqrt {{a^2} + {a^2} + {a^2}} }},\,\frac{a}{{\sqrt {{a^2} + {a^2} + {a^2}} }}} \right\rangle \cr & = \left\langle {\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }}} \right\rangle \cr & AD,\,\left\langle {\frac{a}{{\sqrt {\sum {{a^2}} } }},\,\frac{a}{{\sqrt {\sum {{a^2}} } }},\,\frac{{ - a}}{{\sqrt {\sum {{a^2}} } }}} \right\rangle \cr & = \left\langle {\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }}} \right\rangle \cr & {\text{Angle, }}\theta ,{\text{ between }}AD{\text{ and }}OE{\text{ is given by}} \cr & {\text{cos}}\,\theta = \pm \frac{{\frac{1}{{\sqrt 3 }} \times \frac{1}{{\sqrt 3 }} + \frac{1}{{\sqrt 3 }} \times \frac{1}{{\sqrt 3 }} - \frac{1}{{\sqrt 3 }} \times \frac{1}{{\sqrt 3 }}}}{{\sqrt {\left\{ {{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}} \right\}\left\{ {{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( { - \frac{1}{{\sqrt 3 }}} \right)}^2}} \right\}} }} \cr & = \frac{{\frac{1}{3}}}{{1 \times 1}} = \pm \frac{1}{3} \cr & {\text{Since the cube is in positive octant, we take}}\, + \frac{1}{3} \cr & {\text{So, }}\cos \,\theta = \frac{1}{3} \Rightarrow \theta = {60^ \circ } \cr} $$
[Since value of $$\cos \,\theta $$  decreases as $$\theta $$ increases in $$0$$ to $${90^ \circ }.\cos \,\theta = 1$$   when $$\theta = {0^ \circ }$$  and $$\cos \,\theta = 0$$   when $$\theta = {90^ \circ }$$  ]

In the new frame $$x' = x - {x_1},\,y' = y - {y_1},\,z' = z - {z_1},\,$$
Where $$\left( {{x_1},\,{y_1},\,{z_1}} \right)$$   is shifted origin.
$$ \Rightarrow x' = 0 - 1 = - 1,\,y' = 4 - 2 = 2,\,z' = 5 + 3 = 8$$
Hence, the coordinates of the point with respect to the new coordinates frame are $$\left( { - 1,\,2,\,8} \right).$$