1) we know that any solution to the first two equations is of the form t(-7/3,1,5/3). if this is to be a solution to the third equation as well, we must have:

a(-7/3)t + t - (5/3)t = 0 that is:

t(-7a - 2)/2 = 0.

since it makes no difference if we multiply this by -2 (the RHS is still 0), we can write this as:

t(7a + 2) = 0.

now taking t = 0 works, leading to the solution (0,0,0) which is always a solution of any system of homogeneous equations (ones where the RHS is equal to 0 in every equation).

but if a = -2/7, then the third equation is 0 = 0 no matter what t is, so in this case we have the same solution as with just the first 2 equations: any vector of the form t(-7/3,1,5/3).

(it is worth noting that this is the same set of vectors as the set of vectors of the form s(-7,3,5), since the two vectors in the parentheses only differ by a scalar multiple, so {(-7,3,5)} is a basis for the solution space).

2) here is how i would do the second problem:

i would form the augmented matrix

R_{2}-->-kR_{1}+R_{2}(subtract k times the first row from the second row):

R_{3}-->-4R_{1}+R_{3}:

R_{2}-->15R_{2}

R_{3}-->(-2-3k)R_{3}:

R_{3}-->R_{2}+R_{3}:

just to clean things up, let's divide row 2 by -15:

now let's see what we can do with this. with a little algebraic rearrangement, the last row becomes:

. now IF k ≠ 1, this means that:

<--only defined when k ≠ -25/3.

the second row is now:

IF k ≠ -2/3, we can divide both sides by 2+3k, to get:

<--again, only defined when k ≠ -25/3

the top row now becomes:

, so:

now, i've done the hard stuff, you can decide what happens when k = 1, -2/3 or -25/3.