(2) It should be to find a plane, which can be obtained by the linear combination of two planes that define a straight line, and let a and b be constants:

The plane equation can be expressed as:

a(3x-4y+5z- 10)+b(2x+2y-3z-4)= 0

(3a+2b)x-(4a-2b)y+(5a-3b)z-( 10a+4b)= 0

If the straight line is set to: x=2y=3z=t, then the points on the straight line can be expressed as x=t, y=t/2, z=t/3, and the straight line is parallel to the upper plane. If you substitute the coordinates of these points into the left side of the equation, you should get a constant that is not 0:

(3a+2b) t-(4a-2b) t/2+(5a-3b) t/3-(10a+4b) = (8a/3+2b) t-(10a+4b) = constant,

So 8a/3+2b=0.

8a+6b=0，b=-8a/6=-4a/3，

Recursion: constant =-10a-4 (-4a/3) = a (-10+16/3) =-14a/3.

A is not 0.

Substitute into the previous equation:

(3a-8a/3)x-(4a+8a/3)y+(5a+4a)z-( 10a- 16a/3)= 0

Eliminate one:

x/3-20y/3+9z- 14/3=0

Multiply by 3

x-20y+27z- 14=0

1 1: the distance from point to point on a straight line = √ [(t-2) 2+(t-4) 2+(t-3) 2] = √ (3t2-18t+29) = √ [3 (t)

√ [3 (t-3) 2+2] ≥√2, and the minimum√ 2 is the distance from a point to a straight line.