Why does pattern test not work in this case to restrict the function parameters?

I was looking at this Q&A about using pattern test (pattern_?test) vs pattern condition (pattern_/;cond) and came across this example where using condition was the only possible way to restrict the function parameters (function definition added by me):

Clear[fCond]
fCond[x__ /; Plus[x] == 7] := {x}^2
fCond[1, 2, 4]

(* {1, 4, 16} *)

I know that, aside from the built-in tests in MMA (IntegerQ, NumericQ, eg), one can write custom pattern tests using pure functions [2]:

Hence, I was trying to be clever by using tests with sequence pattern (__) coupled with sequence argument for pure function (##). However, this did not work when plugging in values for the parameter/pattern.

Clear[fTest]
fTest[x__?(Plus[##] == 7 &)] := {x}^2
fTest[1, 2, 4]
(* fTest[1, 2, 4] *)

, even though the pure function test alone does work with the plugged in sequence

Plus[##] == 7 &[1, 2, 4]
(* True *)

Why doesn't pattern test work via this method? If so, is there any other way to do pattern test as opposed to pattern condition in this case?

[2]: Ruskeepaa's Mathematica Navigator 2009 p.498

Answer

My apologies to those who closed this question for my unilaterally reopening it, but there is a nontrivial aspect to this question that I wish to address, and it would not nicely fit in comments. (I am not making an exception for myself; when someone has such an answer he wishs to give to a closed question I nearly always reopen it for him to do so.)

---

While I stated before that: "I can think of no way to do this with PatternTest" was not speaking rigorously and generally, but rather one-to-one substitution of methods. One can apply PatternTest to a larger expression, e.g. the entire left-hand-side, as I did for your prior question. If handling the entire LHS one must exercise care to prevent infinite recursion. Here is an example of how that may be done:

SetAttributes[sumsTo7, HoldFirst]
sumsTo7[_[s__]] := +s == 7

f[x__]?sumsTo7 := {x}

Now:

f[1, 2]
f[3, 4]

f[1, 2]

{3, 4}

It is possible to do this with an anonymous Function as well but it becomes rather baroque:

g[x__]?(Function[Null, Plus @@ Unevaluated[#] == 7, HoldAll]) := {x}

g[1, 2, 3]
g[-3, 10]

g[1, 2, 3]

{-3, 10}